Class 12 Engineering Physics - Dual Nature Of Radiation And Matter

Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (i) red light, and (ii) blue light?

(i) No
(ii) Yes
Reason: According to colour sequence VIBGYOR, the frequency of red light photons is less than threshold frequency of photo metal, but frequency of blue light photons is more than threshold frequency of photo metal; so (i) electrons will not be emitted by red light and (ii) electrons will be emitted by blue light.

A photosensitive surface emits photoelectrons when red light falls on it. Will the surface emit photoelectrons when blue light is incident on it? Give reason.

According to photoelectric equation
$$hv_m=\psi+K.E.$$
$$\Rightarrow \dfrac{hc}{\lambda_{in}}=\psi +K.E.$$ [$$\lambda_{in}$$=Wavelength of incident light]
Since,energy of blue light falling on the surface is more than red light, surface will emit photoelectrons.

Electrons are emitted from the suface when green light is incident on it, but no electrons are ejected when yellow light is incident on it. Do you expect electrons to be ejected when surface is exposed to
(i) Red light and
(ii) Blue light?

(i) The wavelength of red light is longer than threshold wavelength, hence no electron will be emitted with red light.

(ii) The wavelength of blue light is smaller than threshold wavelength, hence electrons will be ejected.

Imagine playing baseball in a universe (not ours!) where the Planck constant is $$0.60 \mathrm{J}$$. s and thus quantum physics affects macroscopic objects. What would be the uncertainty in the position of a $$0.50 \mathrm{kg}$$ baseball that is moving at $$20 \mathrm{m} / \mathrm{s}$$ along an axis if the uncertainty in the speed is $$1.0 \mathrm{m} / \mathrm{s} ?$$

According to the question,

$$ \triangle p = m \triangle v$$

$$ \triangle p = (0.5 kg ) (1 m/s)$$

$$\Rightarrow \triangle p = 0.50 kg m/s$$

$$ \triangle x = \dfrac{h}{ \triangle p} = \dfrac{0.6 J.s}{ 2 \pi (0.5 kg.m/s)} = 0.19 m$$

Calculate the de Broglie wavelength of an electron of mass $$9.11 \times 10^{-31}kg$$ and moving with a velocity of $$1.0 \times 10^6 \,ms^{-1}$$.

Debroglie wavelength of an electron moving with a velocity of $$1 \times 10^6 m /s$$ is given by:

$$\lambda = \dfrac{h}{p} = \dfrac{h}{m_eV} \because \lambda = \dfrac{6.626 \times 10^{-34}}{9.11 \times 10^{-11} \times 1.0 \times 10^6}$$

$$h = 6.626 \times 10^{-34} Js, m_e = 9.11 \times 10^{-31}kg, v = 1.0 \times 10^6 ms^{-1}$$

$$\implies \lambda= 0.7273 \times 10^{-9} = 0.7273 \,nm$$

Define work function for a given metallic surface.

The minimum energy required to free an electron from the metallic surface is called the work function of that surface. it is represented by $$\phi$$

$$\phi=\dfrac{hc}{\lambda_0}$$

where $$\lambda_0$$ = threshold frequency.

How does the maximum kinetic energy of electrons emitted vary with the work function of the metal?

Maximum kinetic energy E_k=hv-W
Clearly, smaller the work function W, greater is the E_k. This means that when work function of a metal increases, maximum kinetic energy of photoelectrons decreases.

How does de Broglie wavelength depend on the momentum of a particle?

$$\lambda=\dfrac{h}{p}\propto \dfrac{1}{p}$$
i.e., de Broglie wavelength varies inversely with the momentum of a particle.

State Heisenberg's uncertainty principle.

2178462_610072_ans_ae82c3b4e25040fca0ae5e58bad08130.jpeg

Calculate the de-Broglie wavelength of the electron orbiting in the $$n=2$$ states of hydrogen atom.

$$n=2$$

$$K.E$$ energy of the second state is:

$$\dfrac { 13.6 }{ { n }^{ 2 } } =\dfrac { 13.6 }{ 4 } $$

$$3.4\times 1.6\times { 10 }^{ -19 }j$$

de brogile wavelength

$$\lambda =\dfrac { h }{ \sqrt { 2m{ E }_{ k } } } $$

$$\dfrac { 6.63\times { 10 }^{ -34 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }\times 3.4\times 1.6\times { 10 }^{ -19 } } } $$

$$=0.67nm$$

Plot the graph between Anode potential and $$\displaystyle\frac{1}{\lambda_{min}}$$. Also find its slope.

The relation between anode potential and minimum wavelength is given by:
$$\displaystyle h\nu = e V_A=\frac{hc}{\lambda}$$, This is same as $$y=mx+c$$ with Slope$$ = \displaystyle \frac{V_A}{1/{\lambda}} = \displaystyle \frac{hc}{e}$$

Define the term work function of a metal.

The work function is the minimum amount of energy (per photon) needed to eject an electron from the surface of a metal

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and
indicate your answer by darkening appropriate bubbles in the 4 4 matrix given in the ORS.

The X-rays contains $$K_{\alpha}, K_{\beta}$$ etc and these are the transitions between two atomic level . So $$A\rightarrow 1$$

Also Hydrogen spectrum has quantized energy levels i.e $$E_n=-\dfrac{136.6}{n^2}$$ and it contains so many transitions between two energy levels like-Lyman, Balmer etc. So $$A\rightarrow 3$$

The photo electric effect is the emission of electron when photon incident on a metal. In beta minus decay, one neutron decay into a proton, an electron and an anti-neutrino. So, So $$B \rightarrow 2, 4$$

Mosley's law is an empirical law concerning the characteristic X-rays. So $$C\rightarrow 1$$

The Einstein's photoelectric equation is $$h\nu=KE_{max}-h\nu_0$$. So $$D \rightarrow 2$$

Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained using Einstein's equation.

(i) For a given metal and frequency of incident radiation, the number of photoelectrons ejected per second is directly proportional to the intensity of incident radiation.

(ii) Maximum kinetic energy of the emitted photo electron is independent of the intensity of incident radiations.

(iii) Emission of photoelectrons is instantaneous.

How can photoelectric effect be used to produce electricity.

Solar cells are grouped into frames called solar panels. These solar cells so arranged in this way, are capable to absorb solar energy from the sun. These solar cells works on the phenomenon of photoelectric effect and hence these solar cells convert solar energy into electrical energy and hence electricity is generated by this way through photoelectric effect.

Give reasons for the following
A 40 watt bulb and 60 watt bulb though have different intensities cannot change the kinetic energy of photoelectrons produced by them.

This is because kinetic energy of emitted photoelectrons in the photoelectric efffect, is independent of intensity of the incoming radiation. So, that's why although 40 watt and 60 watt bulb have different intensities , they cannot change the kinetic energy of photoelecton produced by them.

$$(i)$$ What is thermionic emission?
$$(ii)$$ Name the unit in which the work function of a metal is expressed.

$$(i)$$ The emission of electrons from a metal surface when heat (or thermal) energy is imparted to it is called thermionic emission.
$$(ii)$$ The work function of a metal is expressed in the the unit $$eV$$ (electron volt)

$$1eV=1.6\times {10}^{-19}J$$.

Draw a plot showing the variation of photoelectric current with collector plate potential foe two different frequencies, $$v_1 > v_2$$, of incident radiation having the same intensity. In which case will the stopping potential be higher ? Justify your answer.

The variation of photoelectric current with collector potential for two different frequencies, $$\nu_1>\nu_2$$ is shown in figure.

The photoelectric equation in terms of stopping potential $$V_0$$ is given by , $$eV_0=h\nu-h\nu_o$$

So, $$V_0 \propto \nu$$

Thus, Stopping potential will be higher for $$\nu_1$$.

The following plot is given between $$KE$$ on y-axis and frequency of radiation $$\left(\nu\right)$$ on x-axis. If slope of this plot is $$'m'$$ find ratio $$\left(\displaystyle\frac{m}{h}\right)$$ where $$h = $$ planck's constant.

The Eisenstein's photoelectric equation, $$K_{max}=h\nu-W$$ where W= work function. This is a equation like $$y=mx-c$$

So from the graph, slope, $$m=h$$ and intercept, $$W=428.4$$

Thus, $$m/h=1$$

There are a large number of free electrons moving randomly in a metal. At room temperature, why do they not come out of the surface of metal?

In a metals like copper or tungsten, there are plenty of free electrons. At room temperature these electrons wander randomly in the atomic structure, but they cannot leave the metallic surface. At room temperature ordinary metals do not loose their electrons. This means that a force must exist, which prevents electrons form leaving the metallic surface permanently. To understand what this force is and how it is created, let us assume that due to its random motion, an electron leaves the surface. Immediately after it leaves the surface, the metal gains a positive charge (losing a negatively charged electron is equivalent to gaining a positive charge). This positive charge exerts a force of attraction on the emitted electron. This force pulls the electron back to the metal. For an electron to escape from the metal surface, it must have sufficient kinetic energy to overcome this force. This force is described as surface barrier.

Define the terms (i) 'cut-off voltage' and (ii) 'threshold frequency' in relation to the phenomenon of photoelectric effect.
Using Einstein's photoelectric equations show how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/graph.

i) Cut-off Voltage $$(V_0)$$ is the minimum negative anode voltage for which the photo-current of the circuit reduces to zero.

ii) For a given material, the minimum frequency of the incident frequency below which no emissions of photo-electrons occur, is called threshold frequency $$(\nu_0)$$.

Einstein's equation is $$K_{max}=h\nu-h\nu_0$$

so, $$eV_0=h\nu-h\nu_0$$ or $$V_0=\dfrac{h}{e}(\nu-\nu_0)$$

Now the variation of stopping potential or cut-off potential $$(V_0)$$ with incident frequency $$(\nu)$$ is shown in figure.

We can calculate the threshold frequency from point of intersection on frequency axis and cut-off voltage from point of intersection on potential axis.

(a) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.
(b) Discuss briefly how wave theory of light cannot explain these features
OR
(a) Write the important properties of photons which are used to establish Einstein's photoelectric equation.
(b) Use this equation to explain the concept of (i) threshold frequency of (ii) stopping potential

$$a)$$ The three experimentally observed features in the phenomenon of photoelectric effect are -

i) Threshold frequency: The photoelectric effect will occur when the incident frequency is greater or equal to the threshold frequency for a given metal i.e $$\nu\geq \nu_0$$.

ii) The maximum kinetic energy of photoelectron: When the incident frequency is greater than the threshold frequency, the maximum kinetic energy is proportional to $$(\nu-\nu_0)$$.

iii) Effect of intensity of light : The number of photon incident per unit time per unit area increases with the increase of intensity of incident light. Thus, more the number of photoelectrons increases the photocurrent till it will be saturated.

$$b)$$ We can not explain these by using wave theory of light because there are following reasons-

i) The instantaneous ejection of photoelectrons.

ii) The existence of threshold frequency for a metal surface.

iii) The kinetic energy of photoelectrons is independent of the intensity of light and depends on its frequency.

State two important properties of photon which are used to write Einsteins photoelectric equation. Define (i) stopping potential and (ii) threshold frequency, using Einsteins equation and drawing necessary plot between relevant quantities.

The two important properties of photon for photoelectric effect are

a) The energy of photon is quanta or discrete.

b) The energy of photon is proportional to the frequency of light.

i) Stopping potential : This is the value of negative potential of anode at which the photo electric current in the circuit reduces to zero.

ii) Threshold frequency is the minimum frequency of the incident radiation below which no emission of photoelectric takes place.

The stopping potential $$(V_0)$$ and the threshold frequency $$(f_0)$$ are shown in figure.

The graph shows variation of stopping potential $$V_0$$ versus frequency of incident radiation $$v$$ for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why?

Metal A has higher threshold energy because the minimum frequency required to start photo emission is more in A than that of B.

A monochromatic light source of frequency $$6 \times 10^{14} Hz$$ is emitting energy at the rate $$2 \times 10^{-3} J/s$$. Calculate the number of photons emitted per second by the source.

Let the rate of photons emitted by source be $$n$$.

Energy of single photon=$$h\nu$$

Hence the power of emitted photons=$$P=nh\nu$$

$$\implies n=\dfrac{P}{h\nu}$$

$$=\dfrac{2\times 10^{-3}}{6.626\times 10^{-34}\times 6\times 10^{14}}photons/s$$

$$=5.03\times 10^{15}photons/s$$

Write Einstein's photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation.
The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from $$'\lambda_{1}'$$ to $$'\lambda_{2}'$$. Derive the expressions for the threshold wavelength $$'\lambda_{0}'$$ and work function for the metal surface

Einstein's photoelectric equation states,

$$h\nu=W_0+KE$$

where $$\nu $$ is the frequency of incident light, $$W_0$$ is the work potential of the metal, and $$KE$$ is the maximum kinetc energy of released electron.

The above equations explains the following results:
1. If $$\nu < \nu_o$$, then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if $$\nu > \nu_o$$.
2. The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectron depends only on the frequency of incident light.

From given data,

$$\dfrac{hc}{\lambda_1} = KE + W_o$$

$$\dfrac{hc}{\lambda_2} = 2KE + W_o$$

Eliminating KE, work function is given by:

$$W_o = \dfrac{2hc}{\lambda_1} - \dfrac{hc}{\lambda_2}$$

Threshold wavelength is found by:

$$\dfrac{hc}{\lambda_o} = W_o$$

Using above equations,

$$\lambda_o = \dfrac{\lambda_1 \lambda_2}{2 \lambda_2 - \lambda_1}$$

Write Einstein's photoelectric equations and point out any two characteristic properties of photons on which this equations is based.
Briefly explain the three observed features which can be explained by this caution.

Einstein's photoelectric effect equations:

$$h\nu=\phi_0+K.E.$$
Where $$h \rightarrow$$ Planck's constant
$$\nu \rightarrow$$frequency of the incoming photons
$$\phi_0 \rightarrow$$work functions of the material and
$$KE \rightarrow$$ Kinetic energy
The two characteristic properties of photons on which this equations is based are as follows:
(i) Photons have particle characteristics, It is emitted or absorbed in units called quanta of lights
(ii) Photons have wave characteristic.It travels in space with particular frequency, a characteristic of waves.
Three observed features which can be explained by this equations are:
(i) Solar cells-Also called photo -voltaic cells. It converts solar radiations to electrical emf
(ii) Television telecast -The dark and bright light part of images are interpreted as high and low electrical charges as given by photoelectric emission principle. These are further processed and transmitted
(iii) Burglar alarm- The moment the ultraviolet radiations is cut due to theif, it stops the supply of photons and thus works as "off" mode and the bell rings automatically

Write Einstein's photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation.
Write the three salient features observed in photoelectric effect which can be explained using this equation.

Einstein’s photoelectric equation,

$$K_{max}=h\nu-\phi_0$$ where

$$K_{max}$$ is the max kinetic energy of emitted electrons and $$\phi_0$$ is the work function.

According to Planck’s quantum theory, light radiations consist of small packets of energy. Einstein postulated that a photon of energy hv is absorbed by the electron of the metal surface, then the energy is used to liberate electron from the surface and rest of the energy becomes the kinetic energy of the electron.

Energy of photon is,

$$E=h v$$

Where, h = Planck’s constant

v = frequency of light

The minimum energy required by the electron of a material to escape out of it, is work function.

The additional energy acquired by the electron appears as the maximum kinetic energy ‘K_max’ of the electron.

Einstein's photoelectric equation

$$K_{max}=eV_o$$

Salient features observed in photoelectric effect: —

The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation.

There exists a minimum cut - off frequency , for which the stopping potential is zero.

Photoelectric emission is instantaneous.

What is stopping voltage (or cut voltage)? Plot a graph of variation of photoelectric current with collector plate potential for two incident radiations of same frequency and different intensities.

Photoelectrons are emitted from a metal, when illuminated with light, above the threshold frequency of the metal, with a range of kinetic energies. The stopping potential is the voltage between the metal surface and a cathode, placed close to the surface, in a vacuum, which just stops them from producing a current.

Derive Einstein's photoelectric equation.

Einstein's photoelectric equation.

According to Einstein, the emission of photoelectron is the result of the interaction between a single photon of the incident radiation and an electron in the metal.

When a photon of energy $$hv$$ is incident on a metal surface, its energy is used up in two ways.

(i) A part of the energy of the photon is used in extracting the electron from the surface of metal, since the electrons in the metal are bound to the nucleus. This energy $$W$$ spent in releasing the photoelectron is known as photoelectric work function of the metal. The work function of a photo metal is defined as the minimum amount of energy required to liberate an electron from the metal surface.

(ii) The remaining energy of the photon is used to impart kinetic energy to the liberated electron. If $$m$$ is the mass of an electron and $$v$$ its velocity, then

Energy of the incident photon $$=$$ Work function $$+$$ Kinetic energy of the electron

$$h\gamma = W + \dfrac {1}{2}mv^{2} .... (1)$$

If the electron does not lose energy by internal collisions, as it escapes from the metal, the entire energy $$(h - W)$$ will be exhibited as the kinetic energy of the electron Thus, $$(h - W)$$ represents the maximum kinetic energy of the ejected photoelectron. If Vman is the maximum velocity with which the photoelectron can be ejected, then

$$hv = W + \dfrac {1}{2} mv^{2}_{max} .... (2)$$

This equation is known as Einstein's photoelectric equation.

When the frequency $$(\gamma)$$ of the incident radiation is equal to the threshold frequency $$(\gamma_{0})$$ of the metal surface, kinetic energy of the electron is zero. Then equation $$(2)$$ becomes,

$$h\gamma_{0} = W .... (3)$$

Substituting the value of $$W$$ in equation $$(2)$$ we get,

$$h\gamma - h\gamma_{0} = mv_{max}^{2}$$ (or) $$h(\gamma - \gamma_{0}) = mv_{max}^{2}$$

This is another form of Einstein's photoelectric equation.

The graph between the stopping potential $$V_0$$ and frequency $$v$$ for two different metal plates P and Q are shown in the figure. Which of the metals has greater threshold wavelength and work function?

The threshold frequency of metal plate P is lesser than that of Q.

Since $$\lambda\propto \dfrac{1}{\nu}$$, the threshold wavelength of metal plate P is greater than that of Q.

The work function of a metal is given as

$$W_0=h\nu_0$$

Since the threshold frequency of metal Q is greater, the work function it is also greater than that of P.

State Einstein's photoelectric equation. Explain 'two' characteristics on the basis of this equation.

Einstein's photoelectric equation is $$h\nu=\phi + E_k$$

where $$h$$ is Planck's constant, $$\nu$$ is frequency of electro magnetic radiation, $$h\nu$$$$\rightarrow$$ energy of photon incident on metal surface, $$\phi$$ is the work function and $$E_k$$ is the maximum kinetic energy of the photoelectrons.

Characteristic of photo electric effect:

(a) From the above equation, we find that for photoejection $$h\nu\neq \phi$$ i.e., $$h\nu_{min}=h\nu_0$$ must be equal to $$\phi$$. Hence Photoelectric effect is observed only if $$h\nu\neq h\nu_0$$ or $$v\neq \nu_0$$. This shows the existence of a thershold frequency $$\nu_0$$ for which photo-electrons are just liberated from a metal surface with with $$(KE=0).$$ Since metals differs in electronic configuration the work function $$h\nu_0$$ and therefore, frequency $$\nu_0$$ are different and characteristic of different metals.

(b) From equation $$\displaystyle\frac{1}{2}mv^2_{max}=h\nu-\phi =h(\nu-\nu_0)$$. This shows that the maximum KE increases linearly with the frequency $$\nu$$ of the incident photon $$(\nu\neq \nu_0)$$ and does not depend on the time rate at which photons are incident on a metal surface.

Write any two types of electron emission.

Thermionic emission : - In this type of emission the electron is achieved by heating the electrode. Due to heating the electrons get enough energy that they emit from the surface of that material.

Photoelectric emission : - Electron emission from metallic surface by the application of light is known as photoelectric emission. When a beam of light strikes the surface of cathode normally made of potassium, Sodium the energy of photon of light is transfer to the free electrons of cathode. If the energy of incident photon is greater than the free electrons of the metal then the emission of electron from the metal take place.

Write Einstein's equation of photoelectric effect. Give Einsteins explanation of photoelectric effect.

Einstein's equation of photoelectric effect is

$$E_k=h\nu-\phi=h\dfrac{c}{\lambda}-\phi$$

where $$h=constant,c= \text{speed of light},\lambda=\text{wavelength of light,}\phi={\text{work function of the metal}}$$

On the basis of the Einstein's equation,

When ray of light is incident on the metal surface, then it absorbs this energy and emmits the electron from the surface of the metal.
Some time electron is not ejected because incident energy is less than the work function of the metal.
The maximum kinetic energy of the emerging electrons is independent of intensity of incident radiation but depends linearly on the frequency of the radiation,
Electrons start emitting immediately after the light shines on surface without detectable time delay
For a given frequency of incident radiation, above $$\nu_o$$, the number of electrons emitted per unit time is proportional to the intensity of incident radiation.

Write down Einstein's photoelectric equation.

Einstein's photoelectric equation:
$$E=W_0 + K.E._{max}$$
where $$E$$ is energy of incident photon
$$W_0$$ is work function and
$$K.E._{max}$$ is maximum kinetic energy of photo-electron

What is "work function"?

Work function is a property of a material, which is defined as the minimum quantity of energy which is required to remove an electron to infinity from the surface of a given solid.

State Moseley's law. What is its importance?

Moseley's law states that the square root of the frequency of the emitted x-ray is proportional to the atomic number.
Importance of Moseley's law:

Using this law Moseley arranged K and Ar, Ni and CO in a proper way in Mendeleev's periodic table.
This law was held to the discovery of many new elements like Tc (43), Pr (61), Rh (45).
The atomic number of rare earth have been determined by this law.

What are de-Broglie waves? Establish the de-Broglie wavelength equation.

De-Broglie waves: The waves which are associated with matter are called matter waves or de-Broglie waves.
Wave equation, $$\lambda = \dfrac{h}{mv}$$
where, $$\lambda$$ - wave length, $$h$$ - plank's constant, $$m$$ - mass, $$v$$- velocity of wave.

The planck's constant is $$h$$ and frequency of a photon is $$\nu$$ then write the formula for Einstein photo electric equation.

$$\displaystyle\frac{1}{2}mV^2_{max}=h(\nu-\nu_o)$$.

According to Plank's quantum theory, light is emitted from a source in the forms of bundles of energy called photons. Energy of each photon is $$E=h\nu$$.

Einstein made use of this theory to explain how photo electric emission takes place.

According to Einstein, when photons of energy $$E=h\nu$$ fall on a metal surface, they transfer their energy to the electrons of metal. When the energy of photon is larger than the minimum energy required by the electrons to leave the metal surface, the emission of electrons take place instantaneously.

He proposed that an electron absorbs one whole photon or none. The chance that an electron may absorb more then one electron is negligible because the number of photons is much lower than the electron. After absorbing the photon, an electron either leaves the surface or dissipates its energy within the metal in such a short interval that it has almost no chance to absorb second photon. An increase in intensity of light source simply increases the number of photon and the number of photo electrons but no increase in the energy of photo electron. However, increase in frequency increases the energy of photons and photo electrons.

According to Einstein's explanation of photoelectric emission, a photon of energy $$'E'$$ performs two operations:

1. Removes the electron from the surface of metal

2. Supplies some part of energy to move photo electron towards anode

Since minimum amount of energy to remove electron from a surface is equal to work function, we can write Einstein equation as:

Energy Supplied $$=$$ Energy Consumed in ejecting an electron $$+$$ maximum Kinetic energy of electron

$$E=$$ work function $$+ (KE)_{max}$$

But, $$\phi_o=h\nu_o; E=h\nu and (KE)_{max}=\dfrac{1}{2} mv_o^2$$

So, $$h\nu =h\nu_o+\dfrac{1}{2} mv_o^2$$

$$\implies h(\nu-\nu_o)=\dfrac{1}{2}m v_o^2$$

This is the Einstein's photoelectric equation.

The mass of a moving particle is m and velocity is v, then write the formula for de-Broglie wavelength $$\lambda$$.

When a particle moving with a velocity $$v$$ having a mass of $$m$$ then the formula for de-Broglie wavelength can be written as

$$\lambda = \dfrac{h}{mv}$$, where $$h$$ is plank constant.

Write Einstein's Photo-Electric Equation and explain the terms in it.

When the light means photons incident on a metal surface, the electrons are emitting from metal surface, called photoelectric effect.

The Einstein's Photo-Electric equation : $$KE_{max}=hf-W$$

Here, $$KE_{max}$$ is the maximum kinetic energy of photo-electron, $$h$$ is the Planck's constant, $$f$$ is the frequency of incident light and $$W$$ is the work function of the metal.

Albert Einstein, the great physicist proposed a clear picture to explain photoelectric effect.
a) Explain Einstein's photo electric equation.
b) Name the quanta of light.

a) Einstein realised that it actually takes energy for an electron to leave the surface of the metal. He called this energy the work function, $$\phi$$.

He applied the Conservation of Energy Principle for an electron receiving energy and leaving the metal. Compare energy received and used by the electron at the surface.

Energy supplied to the electron, $$E = hf$$ (energy from the photon)

Energy used by the electron is either: used as work function to escape from the surface or is left with the electron after it has escaped, in which case it is in the form of kinetic energy, $$E_k$$

So for each electron at the surface which leaves the metal:

Energy supplied = energy used

$$hf = \phi + E_k$$

This is Einstein's Photoelectric Equation.

b) According to Plank's quantum theory, light is emitted from a source in the forms of bundles of energy called photons - quanta of light.

Plot a labelled graph of $$|V_s|$$ where $$V_s$$ is stopping potential versus frequency f of the incident radiation.
State how will you use this graph to determine the value of Planck's constant.

Let the work function of the metal be $$\phi$$.

Stopping potential $$V_s = \dfrac{hf}{e} - \dfrac{\phi}{e}$$

Thus the graph of $$|V_s|$$ versus $$f$$ is a straight line having the slope $$\dfrac{h}{e}$$ as shown in the figure.

After finding the slope from the graph, we can determine the value of Plank's constant as Slope $$ = \dfrac{h}{e}$$

$$\implies$$ $$h = e\times $$ Slope

Show that de Broglie wavelength of electrons accelerated $$V$$ volts is very nearly given by:
$$\lambda \left(in \mathring{A} \right) = {\dfrac{150}{V}}^{{1}/{2}}$$.

According to the expression for wavelength:

$$\lambda = \dfrac{h}{\sqrt{2eVm}}$$
$$\lambda = {\left[\dfrac{{h}^{2}}{2eVm} \times {10}^{20}\right]}^{{1}/{2}} \mathring{A}$$
$$={ \left[ \dfrac { { \left( 6.626\times { 10 }^{ -34 } \right) }^{ 2 }\times { 10 }^{ 20 } }{ 2\times 1.6\times { 10 }^{ -19 }\times V\times 9.1\times { 10 }^{ -31 } } \right] }^{ { 1 }/{ 2 } }={ \left[ \dfrac { 150 }{ V } \right] }^{ { 1 }/{ 2 } }$$.

Why do electron not leave the metal surface on their own?

The weaker force of attraction between the nucleus and the valance electrons allows that electron to move freely with in metal surface but this attractive force always attracts the electrons towards the nucleus and so these electrons are bound to the atom and thus they cannot leave the metal surface.

How can electron be made to leave the metal surface?(State any two ways).

If we want to remove the electron from the metal surface, we have to give some energy to the electron which would break the force of interaction between the nucleus and the electron so that it can leave the metal surface.

There are two ways to make the electrons to leave the metal surface :

(1) : Photoelectric effect : A light having energy greater than the work function of the metal is incidented on it and in result the photoelectrons are emitted from the metal surface.

(2) : Thermionic emission : The metal surface is heated which gives sufficient energy to the electrons and so the electron breaks the interaction with the nucleus and leaves the metal surface.

Write Heisenberg's uncertainty principle.

1966507_738519_ans_60188bc541184fa8b993bfe94f781fac.jpg

Explain photoelectric effect on the basis of quantum model and derive the photoelectric equation $$hv=\displaystyle\frac{1}{2}mv^2+hv_0$$, where symbols have their usual meanings.

Einstein gave an explanation for photoelectric emission based on the planks quantum theory of light. When light of certain frequency (greater than threshold frequency) incident on a metallic surface the energy on incident photon (hf) is

Partly used to eject electron from surfaces.

Partly remains as kinetic energy of ejected electrons.

ie, Energy of photon = work function + kinetic energy of electron

Or hf = $$\phi $$+ K. E

Laws of photoelectric emission

There exists a minimum frequency of incident light below which photoelectric emissions is not possible. The frequency is called threshold frequency.

A part of energy of incident electrons in used to eject electrons from a square surface and remaining as kinetic energy of the ejected electrons.

Increase in intensity of the light increase the number of photoelectron.

Increase in frequency of light increase the kinetic energy of electrons.

Derivation of Einstein's photoelectric emission.

Let $$m$$ be the mass of electron and $$V_{max}$$ be the maximum velocity of photoelectron by which it will get ejected.

$$\frac{1}{2}mv_{max}^{2} = hv-\phi_{o}$$

$$\therefore hv = \phi _{o}+\frac{1}{2}mv_{max }^{2}$$

But $$\phi _{o} = hv_{o}$$

$$\therefore hv = hv_{o}+\frac{1}{2}mv_{max}^{2}$$

$$\therefore \frac{1}{2}mv_{max}^{2} = hv - hv_{o}$$

$$\therefore \frac{1}{2}mv_{max}^{2} = h(v-v_{o})$$

This equation is known as Einstein's photoelectric equation.

Explain photoelectric emission on the basis of quantum model of light and derive the equation $$hv=\cfrac{1}{2}m{v}^{2}+h{v}_{0}$$, where symbols used have their usual meanings.

The Quantum Theory of Radiations by Max Planck can explain the photoelectric effect very well. Max Planck says that light is made of bundles of energy known as photons, energy of each photon being equal to $$hv, v$$ being the frequency of the light. So when a photon of frequency being equal to the threshold frequency of the metal strikes the metal surface, it imparts its whole energy to an electron and hence helps the electron to overcome the forces of attraction from the nucleus of the atom. In this way an electron is ejected out of a metal surface when a photon strikes it. If the frequency of the photon is less that $$v_o$$(threshold frequency), then no photoelectric effect is observed.

But if $$v>v_o$$ then some of its energy (which is equal to the ionization energy) is consumed in ejecting the electron from the metal and the remaining is used to impart some kinetic energy to the ejected electron.

$$hv= \phi + \dfrac{1}{2}mv^2$$

Here, $$\phi$$ is the ionization energy, $$v$$ is the velocity imparted to the ejected electron, m is the mass of the electron.

As $$\phi = hv_o$$

$$\therefore hv=\dfrac{1}{2}mv^2+hv_o$$.

Radiations of frequency $$10 ^ { 15 } \mathrm { Hz }$$ are incident on two photosensitive surfaces A and B. Following observations are recorded: Surface A: No photo-emission takes place. Surface $$B :$$ Photo-emission takes place but photo-electrons have zero energy.Explain the above observations on the basis of Einstein's photoelectric equation. How will the observation with a surface $$B$$ change when the wavelength of incitations is decreased?

1868827_1366743_ans_2c18c3223da945c09b0cf2787ca0cdd5.jpg

Explain photoelectric emission on the basis of quantum model of light and derive Einstein's photoelectric emission. Construct logic symbol, Boolean expression and truth table for AND gate.

A B Q
0
0 0
0 1 0
1 0 0
1 1 1
TRUTH TABLE (A. B =Q)
Einstein gave an explanation for photoelectric emission based on the planks quantum theory of light. When light of certain frequency (greater than threshold frequency) incident on a metallic surface the energy on incident photon (hf) is

Partly used to eject electron from surfaces.

Partly remains as kinetic energy of ejected electrons.

ie, Energy of photon = work function + kinetic energy of electron

Or hf = $$\phi $$+ K. E

Laws of photoelectric emission

There exists a minimum frequency of incident light below which photoelectric emissions is not possible. The frequency is called threshold frequency.

A part of energy of incident electrons in used to eject electrons from a square surface and remaining as kinetic energy of the ejected electrons.

Increase in intensity of the light increase the number of photoelectron.

Increase in frequency of light increase the kinetic energy of electrons.

Derivation of Einstein's photoelectric emission.

Let $$m$$ be the mass of electron and $$V_{max}$$ be the maximum velocity of photoelectron by which it will get ejected.

$$\frac{1}{2}mv_{max}^{2} = hv-\phi_{o}$$

$$\therefore hv = \phi _{o}+\frac{1}{2}mv_{max }^{2}$$

But $$\phi _{o} = hv_{o}$$

$$\therefore hv = hv_{o}+\frac{1}{2}mv_{max}^{2}$$

$$\therefore \frac{1}{2}mv_{max}^{2} = hv - hv_{o}$$

$$\therefore \frac{1}{2}mv_{max}^{2} = h(v-v_{o})$$

This equation is known as Einstein's photoelectric equation.

Write Einstein's photoelectric effect and explain lens of photoelectric effect with the help of it.

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The de-Broglie wave length of a particle moving with a velocity $$2.25 \times 10^8 ms^{-1}$$ is equal to the wavelength of a photon. The ratio of kinetic energy of the particle to the energy of the photon is [velocity of light= $$3\times 10^8 ms^{-1}$$]

1322425_1071075_ans_e6e840666aa6413ca9091f8338e79da9.jpg

Stopping potential [V volts] is plotted against frequency of light used $$[\nu]$$. Find work function(eV). $$(h= 6.62\times 10^{-34} J\times s)$$.

We know work = energy and also

Maximum Kinetic energy$$=h\nu \times \Phi $$------(1)

where h= Plank's constant$$=6.62\times { 10 }^{ -34 }\quad\text{ Js}$$

$$\nu =5\times { 10 }^{ 14 }\quad \text{Herts(from figure)}\\ \text{eV}=1.6\times { 10 }^{ -19 }\quad \text{J}$$

Putting the values in 1 we get

Work= energy $$=6.62\times { 10 }^{ -34 }\times (5\times { 10 }^{ 14 })-1.6\times { 10 }^{ -19 }\\ =1.71\times { 10 }^{ -19 }\text{J}$$

The momentum of an object is doubled. How does its kinetic energy change?

The relation between momentum (p) and kinetic energy (E) is : $$E_1= \cfrac{P^2}{2m}$$

Therefore, if P is doubled , $$E_2= \cfrac{4P^2}{2m}$$

Therefore , $$E_2=4E_1$$

Therefore , kinetic energy is quadrapled that is 4 four times the initial one.

Work function of sodium is $$2.3 eV$$. Does sodium show photo-electric emission for light wavelength $$6800 \mathring {A}$$?

from the photoelectric equation.

$$E=W+K.E$$

$$E=hv=\dfrac{1240}{680}=1.82eV$$

which is less than $$2.3eV$$

no photoelectric effect would happen.

The equation for a wave travelling in x-direction on a string is : y = 3(sin3.14x-314t),then
Find the maximum velocity of particle of the string .

Equation of travelling wave is $$y=3\sin \left[ (3.14c{{m}^{-1}}x)-(314{{s}^{-1}}t) \right]$$

From the above equation,

$$ \omega =314 $$

$$ 2\pi \nu =314 $$

$$ \nu =50\,Hz $$

$$ k=3.14c{{m}^{-1}}=314\,{{m}^{-1}} $$

$$ \dfrac{2\pi }{\lambda }=314 $$

$$ \lambda =\dfrac{2}{100}\,m=\dfrac{1}{50}\,m $$

So, velocity

$$v= \nu \lambda$$

$$ v=50\times \dfrac{1}{50} $$

$$ v=1\,m/s $$

What should be the ratio of velocities of $$CH_4$$ and $$O_2$$ molecules so that they are associated with de Brogile waves of equal wave lenghs ?
According to be Broglie equation = $$\lambda = \dfrac{h}{mv}$$

$$\lambda_{CH_4} = \dfrac{h}{(mv)_{CH_4}} = \dfrac{h}{16 \times v_{CH_4}} $$
$$\lambda_{O_2} = \dfrac{h}{(mv)_{O_2}} = \dfrac{h}{32 vo_{O_2}}$$
Since $$> CH_4 = > O_2$$
$$\Rightarrow \dfrac{h}{16 v_{CH_4}} = \dfrac{h}{32 v_{O_2}} \Rightarrow \dfrac{v_{CH_4}}{v_{O_2}} = \dfrac{32}{16} = \dfrac{2}{1}$$

Mention any two types of electron emission.

· Field Emission: In this type, a very strong electric field is applied to the metal which pulls the electrons out of the surface due to the attraction of the positive field.

· Photoelectric Emission: In this type, the light of a certain frequency is made to fall on the metal surface which leads to the emission of electrons.

A particle of mass M at rest decays into two particles of masses $$m_1$$ and $$m_2$$ having non zero velocities. The ratio of the de Broglie wavelengths of the particles, $$\lambda_1/\lambda_2$$ is?

De Broglie wavelengths is given by $$\lambda =\frac { h }{ p } $$

where $$h$$ is planks constant and $$p$$ is momentum

Now mass M at rest initially from the momentum of conservation

$${ p }_{ 1 }={ p }_{ 2 }$$

Since the momentum is equal, the de broglie wavelength is equal

Ans $$1$$

Which colored light bulb - red, orange, yellow, green, or blue- emits photons with (a) the least energy and (b) the greatest energy? Account for your answers.

$$(a)\ $$ Red-colour light bulb emit photons with the lowest frequency compared to light bulbs of other colours.

$$\therefore \ $$ Red-colour light bulb emits photons with lowest energy.

$$(b)\ $$ Colour Blue appears next to violet in the continuous visible spectrum $$\therefore \nu_{Blue} < \nu_{violet}$$ but greater than other colours given here.

Hence, Blue- Colour light bulb emits photons with greater energy.

In the arrangement shown in figure , y=1.0mm, d=0.24mm and D= 1.2m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.

Given :

The fringe width is $$y=1.0\times 2\ mm = 2.0 mm$$

The distance between the slits is $$ d = 0.24 mm$$

The work function of the material is $$W_0=2.2eV$$

The distance of the screen from the slits is $$D=1.2\ m$$

The fringe width is given as:
$$y=\dfrac { \lambda D }{ d } $$
or, $$\lambda =\dfrac { yd }{ D } =\dfrac { 2\times { 10 }^{ -3 }\times 0.24\times 10^{ -3 } }{ 1.2 } =4\times 10^{ -7 }m$$

The energy of the photon is obtained as
:

$$E=\dfrac { hc }{ \lambda } =\dfrac { 4.14\times 10^{ -15 }\times 3\times { 10 }^{ 8 } }{ 4\times 10 } =3.105\quad eV$$

Stopping potential $$eV_0=3.105-2.2=0.905V$$

Show that de-Broglie wavelength of electron accelerated through V volt is nearly given by:
$$\lambda=[\frac{1550}{V}]^{1/2}$$ in $$A^{\circ}$$

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An electron in hydrogen like atom, is in an excited state. It has a total energy of -3.4 eV. Calculate.
(a)The kinetic energy of electron
(b) The de broglie wavelength of electron
$$(h=6.6 \times 10^{-34}, m_e= 9.108 \times 10^{-31}kg)$$

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You wish to pick an element for a photocell that will operate via the photoelectric effect with visible light. Which of the following are suitable (work functions are in parentheses): tantalumeV $$),$$ tungsten $$(4.5 \mathrm{eV}),$$ aluminum $$(4.2 \mathrm{eV}),$$ barium $$(2.5 \mathrm{eV})$$ lithium $$(2.3 \mathrm{eV}) ?$$

The energy of the most energetic photon in the visible light range (with wavelength of about $$400 \mathrm{nm}$$ ) is about $$E=(1240 \mathrm{eV} \cdot \mathrm{nm} / 400 \mathrm{nm})=3.1 \mathrm{eV}$$ (using the value $$h c=1240 \mathrm{eV} \cdot \mathrm{nm}$$ ). Consequently, barium and lithium can be used, since their work functions are both lower than $$3.1 \mathrm{eV}$$

A nonrelativistic particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $$1.813 \times 10^{-4}$$. By calculating its mass, identify the particle.

For $$U=U_{0},$$ Schrödinger's equation becomes

$$\dfrac{d^{2} \psi}{d x^{2}}+\dfrac{8 \pi^{2} m}{h^{2}}\left[E-U_{0}\right] \psi=0\\$$

We substitute $$\psi=\psi_{0} e^{i k x} .$$ The second derivative is

$$\dfrac{d^{2} \psi}{d x^{2}}=-k^{2} \psi_{0} e^{i k x}=-k^{2} \psi\\$$

The result is

$$-k^{2} \psi+\dfrac{8 \pi^{2} m}{h^{2}}\left[E-U_{0}\right] \psi=0\\$$

Solving for k, we obtain

$$k=\sqrt{\dfrac{8 \pi^{2} m}{h^{2}}\left[E-U_{0}\right]}=\dfrac{2 \pi}{h} \sqrt{2 m\left[E-U_{0}\right]}\\$$

Write the wave function $$\psi(x)$$ displayed in Eq. $$38-27$$ in the form $$\psi(x)=a+i b,$$ where $$a$$ and $$b$$ are real quantities. (Assume(b) Write the time-dependent wave function $$\Psi(x, t)$$ that $$\psi_{0}$$ is real. that corresponds to $$\psi(x)$$ written in this form.

If the momentum is measured at the same time as the position, then

$$\Delta p \approx \dfrac{\hbar}{\Delta X}=\dfrac{6.63 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}}{2 \pi(50 \mathrm{pm})}=2.1 \times 10^{-24} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\\$$

The uncertainty in the position of an electron along an $$x$$ axis is given as $$50 \mathrm{pm},$$ which is about equal to the radius of a hydrogen atom. What is the least uncertainty in any simultaneous measurement of the momentum component $$p_{x}$$ of this electron?

(a) The transmission coefficient $$T$$ for a particle of mass $$m$$ and energy $$E$$ that is incident on a barrier of height $$U_{b}$$ and width $$L$$ is given by

$$T=e^{-2 b L}\\$$

where

$$b=\sqrt{\dfrac{8 \pi^{2} m\left(U_{b}-E\right)}{h^{2}}}\\$$

For the proton, we have

$$\begin{array}{l}b=\sqrt{\dfrac{8 \pi^{2}\left(1.6726 \times 10^{-27} \mathrm{kg}\right)(10 \mathrm{MeV}-3.0 \mathrm{MeV})\left(1.6022 \times 10^{-13} \mathrm{J} / \mathrm{MeV}\right)}{\left(6.6261 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\right)^{2}}} \\=5.8082 \times 10^{14} \mathrm{m}^{-1}\end{array}\\$$

This gives $$b L=\left(5.8082 \times 10^{14} \mathrm{m}^{-1}\right)\left(10 \times 10^{-15} \mathrm{m}\right)=5.8082,$$ and

$$T=e^{-2(5.8082)}=9.02 \times 10^{-6}\\$$

The value of $$b$$ was computed to a greater number of significant digits than usual because an exponential is quite sensitive to the value of the exponent.

(b) Mechanical energy is conserved. Before the proton reaches the barrier, it has a kinetic energy of $$3.0 \mathrm{MeV}$$ and a potential energy of zero. After passing through the barrier, the proton again has a potential energy of zero, thus a kinetic energy of $$3.0 \mathrm{MeV}$$

(c) Energy is also conserved for the reflection process. After reflection, the proton has a potential energy of zero, and thus a kinetic energy of $$3.0 \mathrm{MeV}$$

(d) The mass of a deuteron is $$2.0141 \mathrm{u}=3.3454 \times 10^{-27} \mathrm{kg},$$ so

$$\begin{array}{l}b=\sqrt{8 \pi^{2}\left(3.3454 \times 10^{-27} \mathrm{kg}\right)(10 \mathrm{MeV}-3.0 \mathrm{MeV})\left(1.6022 \times 10^{-13} \mathrm{J} / \mathrm{MeV}\right)} \\=8.2143 \times 10^{14} \mathrm{m}^{-1}\end{array}\\$$

This gives $$b L=\left(8.2143 \times 10^{14} \mathrm{m}^{-1}\right)\left(10 \times 10^{-15} \mathrm{m}\right)=8.2143,$$ and

$$T=e^{-2(8.2143)}=7.33 \times 10^{-8}\\$$

(e) As in the case of a proton, mechanical energy is conserved. Before the deuteron reaches the barrier, it has a kinetic energy of $$3.0 \mathrm{MeV}$$ and a potential energy of zero. After passing through the barrier, the deuteron again has a potential energy of zero, thus a kinetic energy of $$3.0 \mathrm{MeV}$$

(f) Energy is also conserved for the reflection process. After reflection, the deuteron has a potential energy of zero, and thus a kinetic energy of $$3.0 \mathrm{MeV}$$

The frequency v of incident radiation is greater that threshold frequency $$v_0$$ in a photocell. How will the stopping potential vary if frequency (v) is increased, keeping other factors constant?

From Einstein's photoelectric equation, stopping potential $$V_0$$ is
$$eV_0=hv-hv_0\Rightarrow V_0=\dfrac{h}{e}(v-v_0)$$
Given $$v>v_0$$, so with increase of frequency v, stopping potential increases.

Two metals A and B have work functions 4eV and 10eV respectively. Which metal has the higher threshold wavelength?

Work function $$W=hv_0=\dfrac{hc}{\lambda_0}$$
$$\Rightarrow \lambda_0\propto \dfrac{1}{W}$$
As $$W_A<W_B;(\lambda_0)_A>(\lambda_0)_B$$
i.e., threshold wavelength of metal A is higher.

Write the basic features of photon picture of electromagnetic radiation on which Einstein's photoelectric equation is based.

Features of the photons:
(i) Photon are particles of light having energy $$E=hv$$ and momentum $$p=\dfrac{h}{\lambda}$$, where h is Planck constant.
(ii) Photons travel with the speed of light in vacuum, independent of the frame of reference.
(iii) Intensity of light depends on the number of photons crossing unit area in a unit time.

If the intensity of radiation in a photocell is increased how does the stopping potential vary?

The stopping potential does not depend on the intensity of incident radiation; so stopping potential will remain unchanged.

it only depends on the frequency of the incident ray.

Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies $$V_A>V_B$$
(i) In which case is the stopping potential more and why?
(ii) Does the slope of the graph depend on the nature of the material used? Explain

(i) From the graph for the same value of 'v', stopping potential is more for material 'B'
From Einstein's photoelectric equation
$$eV_0=hv-hv_0$$

$$V_0 =\dfrac{h}{e}-\dfrac{h}{e}_0=\dfrac{h}{e}(v-v_0)$$

Therefore$$ V_0$$ is higher for lower value of $$V_0$$

ii. No, as slope is given by $$\dfrac{h}{e}$$ which is a universal constant.

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Two lines A and B shown in the graph represent the de Broglie wavelength$$(\lambda)$$ as a function of $$\dfrac{1}{\sqrt{V}}$$(V is the accelerating potential) for two particles having the same charge. Which of the two represents the particle of smaller mass?

De Broglie wavelength $$\lambda=\dfrac{h}{\sqrt{2mqV}}$$ or $$\lambda=\dfrac{h}{\sqrt{2mq}}$$.$$\dfrac{1}{\sqrt{V}}$$
The graph of $$\lambda$$ versus $$\dfrac{1}{\sqrt{V}}$$ is a straight line of slope $$\dfrac{h}{\sqrt{2mq}}\propto \dfrac{1}{\sqrt{m}}$$. The slope of line B is large, so particle B has smaller mass.

Chlorophyll present in green leaves of plants absorbs light at $$4.620\times 10^{14}Hz$$. Calculate the wavelength of radiation in nanometer, Which part of the electromagneitc spectrum does it belong to?

$$\lambda =\dfrac{c}{v}=\dfrac{3.0\times 10^{8}m/s}{4.620\times 10^{14}Hz}$$
$$=0.6494\times 10^{-5}m=649.4\ nm$$
It belongs to the visible light of the spectrum.

According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. How ever, a cricket ball of mass $$100\ g$$ does not move like a wave when it is thrown by a bowler at a speed of $$100\ km/h$$. Calculate the wavelength of the ball and explain why it does not show wave nature.

According to be Broglie.
The wavelength

λ=hmvλ=hmv
Since the wavelength is too small to be detected so, it does not show wave nature.

In the stydy of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:
(i) Which one of the two metals has higher threshold frequency?
(ii) Determine the work function of the metal which has greater value.
(iii) Find the maximum kinetic energy of electron emitted by light of frequency $$8\times10^{14}Hz$$ for this metal.

(i) Threshold frequency of P is $$3\times10^{14}Hz.$$
Threshold frequency of Q is $$6\times10^{14}Hz.$$
Clearly Q has higher threshold frequency.
(ii) Work function of metal Q,$$\psi_0=hv_0$$
$$=(6.6\times10^{-34})\times6\times10^{14}J$$
$$=\dfrac{39.6\times10^{-20}}{1.6\times10^{-19}}eV=2.5eV$$
(iii) Maximum kinetic energy, $$K_{max}=hv-hv_0$$
$$=h(v-v_0)$$
$$=6.6\times10^{-34}(8\times10^{14}-6\times10^{14})$$
$$=6.6\times10^{-34}\times2\times10^{14}J$$
$$=\dfrac{6.6\times10^{-34}\times2\times10^{14}J}{1.6\times10^{-19}}eV$$
$$\therefore K_{max}=0.83eV$$

Wavelength of different radiations are given below:
$$\lambda=(A)300\ nm$$
$$\lambda (B)=300\mu\ m$$
$$\lambda (C)=3\ nm$$
$$\lambda (D)=30\overset { o }{ A } $$
Arrange these radiations in the increasing order of their energies.

$$E\alpha\dfrac{1}{\lambda}$$
$$300\mu m < 300\ nm < 2\ nm= < 30\overset { o }{ A } $$
$$\lambda (B) < (A) < \lambda (C)=\lambda (D)$$

Assuming an electron is confined to a 1nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle $$(\Delta x \times \Delta p \cong h)$$. You can assume the uncertainty in position $$\Delta x$$ as 1nm. Assuming $$p \cong \Delta p$$, find the energy of the electron in electron volts.

As electron revolves in circular path so $$\Delta r = 1 \ nm= 10^{-9} m$$

$$\Delta x \times \Delta p \cong h$$

$$\Delta p = \dfrac{h}{\Delta x} = \dfrac{6.62 \times 10^{-34}}{2 \pi \Delta r}Js$$

$$= 3.8 \times 10^{-2} \ eV$$

Explain the working of photo emissive cell. Write any two applications of photoelectric cells.

A photo-emissive cell is a type of gas-filled or vaccum tube that is sensitive to light.

It makes use of photo electric effects, the phenomenon whereby light sensitive surface give off electrons when stuck by light.

when photons strike the cathod the electrons in the ligh-sensitive substance absorb the energy. The electrons thus aquire enough energy to escape from the surface of the metal. So, when potential difference is applied, electrons move towards anode creating an electric current in opposite direction.

Applications

i) controlling temperature of furnace

ii) Automatic switching on and off of street bulbs

iii) Reproducing sound in cinematography.

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A hypothetical electromagnetic wave is shown in Fig $$2.2$$. Find out the wavelength of the radiation.

Wavelength: It may be defined as the distance between two neighboring crest or troughs of a wave as shown. It is denoted by $$\lambda$$.
$$\lambda =4\times 2.16\ pm=8.64\ pm$$

Write Heisenberg Uncertainty principle in relation to position and momentum of particle.

Heisenberg Uncertainty principle in relation to position and momentum of particle is $$ \Delta x \Delta p \geq \dfrac{h}{2 \pi} $$

The uncertainty in the X-component of momentum of a moving electrons is $$ 13.18 \times 10^{-30}$$ kg-m/s. Find out the uncertaintities in the X-component of position and velocity.

$$ \Delta p_x = 13.18 \times 10^{-30}$$ kg m/s

Position $$ (\Delta x) = $$ ?

$$ \Delta p_{x} \Delta x \geq \dfrac{h}{2} $$ or $$ \Delta p_x \times \Delta x = \dfrac{h}{4 \pi} $$

$$ \Delta x = \dfrac{h}{4 \pi \times \Delta p_x} = \dfrac{6.63 \times 10^{-34}}{4 \times 3.14 \times 13.18 \times 10^{-30}} $$

$$ = 0.040 \times 10^{-4} $$

$$ = 0.40 \times 10^{-5}\,m $$

For component of velocity

$$ \Delta p_{x} = mv_{x} $$

$$ \therefore $$ $$ \Delta v_x = \dfrac{\Delta p_x}{m} = \dfrac{13.18 \times 10^{-30}}{9.1 \times 10^{-31}} $$

$$ = 1.448 \times 10 = 14.48 \, m/s $$

What explanation was given by Einstein to explain photoelectric effect ? What do you understand by threshold frequency ?

Einstein’s Photoelectric Equation and Explanation of Experimental Results of Photoelectric Effect on the Basis of this Equation

In $$1905$$, Albert Einstein $$(1879-1955)$$ proposed a radically new picture of electromagnetic radiation to explain photoelectric effect. In this picture, photoelectric emission does not take place by continuous absorption of energy from radiation. Radiation of energy is built up of discrete units : the so called quanta of energy of radiation. Each quanta of radiant energy has energy $$hv$$, where $$h$$ is Planck’s constant and $$v$$ the frequency of light. In photoelectric effect, an electron absorbs a quanta of energy $$(hv)$$ of radiation. If the energy of quanta absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function $$ \phi_0 $$), the light energy of radiation is given to the metal surface. This energy is spent into two ways. One part of energy is required to just eject the electron from metal surface and rest part of energy to provide the kinetic energy to the photon. So by the law of conservation of energy,

The energy of light = Work function + Kinetic energy

$$ hv = \phi + K_{max} $$ ..........(1)

When $$ v_{max} = 0 \Rightarrow \, K_{max} = 0 $$

then $$ v = v_0 $$

$$ \therefore \, hv_0 = \phi + 0 $$

$$ \therefore\, hv_0 = \phi $$ ...........(2)

$$ \therefore \, hv = hv_0 + K_{max} $$ ...........(3)

Equation (3) is called Einstein's photoelectric equation, basically it is a statement of law of conservation of energy for absorption of a single photon by a metal of work function $$ \phi_0 $$.

If mass of emitted electrons is $$m$$ and maximum velocity is $$v_{max} $$ then

$$ hv = \dfrac{1}{2} mv^{2}_{max} + \phi_{0} $$ ................(4)

If stopping potential is $$ V_0 , K_{max} = eV_0$$

$$ hv = eV_0 + \phi_0 $$ ..............(5)

Equations (4) and (5) are the other forms of photoelectric equation. Now we see how this equation accounts in a simple and elegant manner for all the observations on photoelectric effect.

(i) According to equation (3), $$ K_{max} $$ depends linearly on frequency of incident radiation (v), and is independent of intensity of radiation, in the agreement with observation. This has happened because in Einstein’s picture, photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. The intensity of radiation (that is proportional to the number of energy quanta per unit area per unit time) is irrelevant to this basic process.

(ii) Since $$ K_{max} $$ must be non-negative, equation (3) implies that photoelectric emission is possible only if

$$ hv > \phi_0$$

or $$ hv > hv_0 $$ , where

$$ v_0 = \dfrac{\phi_0}{h} $$ .........(6)

Equation (6) shows that the greater the work function $$ \phi_0$$ , the higher the minimum or threshold frequency $$ v_0$$ needed to emit photoelectrons. Thus there exists a threshold frequency below which no photoelectric emission is possible, no matter how intense the incident radiation is or how large it falls on the surface.

(iii) In this picture, intensity of radiation as noted above, is proportional to the number of energy of quanta per unit area per unit time. The greater the number of energy quanta available, the greater is the number of electrons absorbing the energy quanta and greater, therefore, is the number of electrons coming out from the metal (for $$ v > v_0$$). This explains why, for $$ v > v_0 $$ , photoelectric current is proportional to intensity.

(iv) In Einstein’s picture, the basic elementary process involved in photoelectric effect is the absorption of light quantum by an electron. This process is instantaneous. Thus whatever may be the intensity i.e. the number of quanta of radiation per unit area per unit time, photoelectric emission is instantaneous. Intensity only determines how many electrons are able to participate in the elementary process (absorption of a light quantum by a single electron), and therefore, the photoelectric current.

$$ \because \, K_{max} = eV_0 $$, so the photoelectric equation (3) can be written as

$$ eV_0 = hv - \phi_0 $$ ; for $$ v \geq v_0 $$

or $$ V_0 = \left ( \dfrac{h}{e} \right ) v - \dfrac{\phi_0}{e} $$ ......(7)

This is an important result. It predicts that the $$V_0$$ versus $$v$$ curve is a straight line with slope = $$ (h/e) $$, independent of the nature of the material. During $$1906 -1916$$, Millikan performed a series of experiments on photoelectric effect, aimed at disproving Einstein’s photoelectric equation. He measured the slope of the straight line obtained for sodium, similar to that shown in figure. Using the known value of $$e$$, he determined the value of Planck’s constant $$h$$. This value was close to the value of Planck’s constant $$ (= 6.626 \times 10^{-34}\, J-s) $$ determined in an entirely different reference. In this way, in $$1916$$, Millikan traced the validity of Einstein’s photoelectric equation instead of disproving it.

The successfully explanation of photoelectric effect using the hypothesis of light energy or quanta and the experimental determination of values of $$h$$ and in agreement with values obtained from other experiments, led to the acceptance of Einstein’s picture of photoelectric effect. Millikan verified photoelectric equation with great precision, for a number of alkali metals over a wide range of radiation frequencies.

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The time period of electromagnetic pulse is $$0.30\,ms$$. Find out the uncertaintity in the energy of photon.

$$ \Delta t = 0.30 \times 10^{-3} s ; E = $$ ? , $$ \therefore \, \Delta E \times \Delta t \geq \dfrac{h}{2} $$

or $$ \Delta E \times \Delta t = \dfrac{h}{4 \pi} $$

$$ \Delta E = \dfrac{6.63 \times 10^{-34}}{4 \times 3.14 \times 0.30 \times 10^{-3}} $$

$$ = 1.76 \times 10^{-31} \,J $$

A proton with KE $$6MeV$$ is travelling in a particle accelerator of circular orbit $$0.75m$$. What fraction of energy does it radiate per sec?

Using Larmor's Formula

$$ P=\dfrac { 2k{ q }^{ 2 }{ a }^{ 2 } }{ 3{ c }^{ 3 } } \quad where,\\ P=\quad Power\quad radiated,\\ k=\quad Coulomb's\quad constant,\\ a=acceleration\quad of\quad particle,\\ c=\quad speed\quad of\quad light\\ now\quad a=\dfrac { 2K }{ mr } ,\quad where\\ K=\quad Kinetic\quad energy,\\ r=radius\quad of\quad circle\\ \\ now\quad putting\quad it\quad in\quad the\quad equation\quad we\quad get\\ \frac { P }{ K } =\dfrac { 8k{ q }^{ 2 }K }{ 3{ m }^{ 2 }{ c }^{ 3 } } $$

on solving and putting values in correct units we get,

$$ fraction\quad (\dfrac { P }{ K } )=0.047 $$

The electron energy in hydrogen atom is given by $$E_n$$ = $$\dfrac{-21.7*10^{-19}}{n^2}$$ergs. Calculate the energy required to remove an $$e^-$$ completely from n = 2 orbit. What is the largest wavelength in cm of light that can be used to cause this transition?

$${ E }_{ 2 }=\dfrac { -21.7\times { 10 }^{ -19 } }{ 4 } ergs=-5.425\times { 10 }^{ -26 }J\\ \quad { E }_{ 2 }=\dfrac { hc }{ \lambda } \Rightarrow 5.425\times { 10 }^{ -26 }=\dfrac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ \lambda } \\ \lambda =\dfrac { 6.626\times 3 }{ 5.425 } m\\ \lambda =3.66m=366cm$$

An electron moves in an electric field with a kinetic energy of $$2.5 eV$$. What is the associated de Broglie wavelength?

Kinetic energy $$=\dfrac { 1 }{ 2 } m{ v }^{ 2 }$$ $$\left( v=\dfrac { h }{ m\lambda } \right) $$

$$=\dfrac { 1 }{ 2 } m{ \left( \dfrac { h }{ m\lambda } \right) }^{ 2 }$$

$$=\dfrac { 1 }{ 2 } \dfrac { { h }^{ 2 } }{ m{ \lambda }^{ 2 } }$$

or $${ \lambda }^{ 2 }=\dfrac { 1 }{ 2 } \dfrac { { h }^{ 2 } }{ m\times KE }$$

$$\lambda =\dfrac { h }{ \sqrt { 2m\times KE } }$$ $$\begin{pmatrix} m=9.108\times { 10 }^{ -28 }g \\ h=6.626\times { 10 }^{ -27 }erg-sec \\ 1eV=1.602\times { 10 }^{ -12 }erg \end{pmatrix}$$

$$=\dfrac { 6.626\times { 10 }^{ -27 } }{ \sqrt { 2\times 9.108\times { 10 }^{ -28 }\times 2.5\times 1.602\times { 10 }^{ -12 } } } $$

$$=7.7\times { 10 }^{ -8 } cm$$.

Find the de-Broglie wave length of neutron at $$ {127^\circ }C$$.

$$T= 127+273=400K$$

De Broglie wavelength of thermal neutron,

$$\lambda= \cfrac {30.835 \overset {o}{A}}{\sqrt {T}}=\cfrac {30.8 \overset {o}{A}}{\sqrt {400}}= 1.54 \overset {o}{A}$$

A bulb emits light of wavelength $$4500\mathring{A}$$, the bulb is rated as $$150$$ Watt and $$8$$% of the energy is emitted as light. How many photons are emitted by the bulb per second?

Electrical energy in joules = Power in watts $$ \times $$ Time in seconds

Thus $$150$$ watt = $$150$$ joules of energy emitted per second.

$$ \therefore $$ Energy emitted as light

$$ = \dfrac{8}{100} \times 150 = 12 J $$

$$ E = n h \nu = nh \dfrac{c}{\lambda} $$

$$ \therefore $$ $$ n = \dfrac{E \times \lambda}{h \times c} $$

$$ = \dfrac{(12\ J) \times (4500 \times 10^{-10} m)}{(6.626 \times 10^{-34} Js ) \times (3 \times 10^{8} \, ms^{-1})} $$

$$ = 2.717 \times 10^{19} $$

$${{\lambda }}_{\text{m}}^{\text{o}}\,$$ for $${\text{NaCl,}}\,\,{\text{HCl}}\,{\text{and}}\,{\text{NaAc}}$$ are 126.4, 425.9 and 91.0 S $${\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{ - 1}}$$ respectively. Calculate $${{{\lambda }}^{\text{o}}}$$ for HAc.

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Draw a graph showing the variation of stopping potential $$(V_{0})$$ with the $$3$$ frequency of incident radiation $$(v)$$ in relation to the electric effect.
How can the value of Plancks constant be determined from this graph?

Slope of the graph corresponds to $$\frac{h}{e}$$ which is a

constant and independent of the material.

so tan $$\phi$$ = $$\frac{h}{e}$$

therefore

h = $$\frac{e}{tan \phi}$$

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Draw a graph showing the variation of stopping potential with frequency of incident radiation for two photosensitive materials, having work function $${ W }_{ 1 } $$ and $${ W }_{ 2}$$ $$({ W }_{ 1 }>{ W }_{ 2 })$$. Write the two important conclusion that can be drawn from the study of these plots.

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Consider a thin target ($$10^{-2}\ m$$ square, $$10^{-3}\ m$$ thickness ) of sodium, which produces a photo-current of $$100\ \mu A$$ when a light of intensity $$100\ W/m^{2}(\lambda=660\ nm$$) falls on it. Find the probability that a photo-electron is produced when a photon strikes a sodium atom.
[ Take density of $$Na=0.97\ kg/m^{3}$$ ]

$$6\times{10}^{26}$$Na atoms weighs $$23$$kg

Volume of target$$={10}^{-4}\times{10}^{-3}={10}^{-7}{m}^{3}$$

Density of sodium$$=d=0.97$$kgper cubic metre.

Volume of $$6\times{10}^{26}$$ Na atoms$$=\dfrac{23}{0.97}=23.7{m}^{3}$$

Volume occupied of $$1$$ Na atom

$$=\dfrac{23}{0.97\times6\times{10}^{26}}{m}^{3}=3.95\times{10}^{-26}{m}^{3}$$

Number of sodium atoms in the target$$=\dfrac{{10}^{-7}}{3.95\times{10}^{-26}}=2.53\times{10}^{18}$$

Number of photons in the beam for $${10}^{-4}{m}^{2}=n$$

Energy per $$s$$ $$nh\nu={10}^{-4}$$J$$\times100={10}^{-2}$$W

$$h\nu$$ for $$\lambda=660$$nm$$=\dfrac{1234.5}{600}$$

$$2.05eV=2.05\times 1.6\times{10}^{-19}=3.28\times {10}^{-19}$$J

$$n=\dfrac{{10}^{-12}}{3.28\times{10}^{-19}}=3.05\times{10}^{16}/s$$

$$n=\dfrac{1}{3.2}\times{10}^{17}=3.1\times{10}^{16}$$

If $$P$$ is the probability of emission per atom, per photon, the number of photoelectrons emitted/second.

$$=P\times 3.1\times{10}^{16}\times 2.53\times{10}^{18}$$

Current$$=P\times 3.1\times{10}^{16}\times 2.53\times{10}^{18}\times 1.6\times{10}^{-19}$$

$$=P\times 1.25\times{10}^{16}$$A

This must be equal to $$100\mu $$A

or $$P=\dfrac{100\times{10}^{-6}}{1.25\times{10}^{16}}=8\times{10}^{-21}$$

Thus the probability of photoemission by a single photon on a single atom is very much less than $$1.$$ (That is why absorption of two photons by an atom is negligible).

Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions $$W _ { 1 }$$ and $$W _ { 2 } \left( W _ { 1 } > W _ { 2 } \right) .$$ On what factors does the (i) slope and (ii) intercept of the line depend?

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Using photon picture of light, show how Einstein's photoelectric equation can be established. Write two features of photoelectric effect which cannot be explained by wave theory.

Light consists of photons. When a photon inferacts with an electron it gives its entire energy to the electron and then exists no longer. Energy used to knock out the electron. (hv) is work function and rest energy is given to electron is kE.

(i) Greater intensity has no effect on kE of an electron cannot be explained by wave theory.

(ii) Wave theory also fails to explain the existence of threshold energy.

Calculate the wavelength of moving electron having $$ 4.55 \times 10^{-25}$$ J of kinetic energy.

Kinetic energy $$=\dfrac {1}{2} mu^2= 4.55 \times 10^{-25}$$J

$$\therefore u^2= \dfrac {2\times 4.55\times 10^{-25}}{9.108 \times 10^{-31}}$$

$$\Rightarrow u=999.6\ m/s\approx 10^3\ m sec^{-1}$$

$$\therefore \lambda = \dfrac {h}{mu}=\dfrac{6.626\times 10^{-34}}{9.108 \times 10^{-31}\times 10^3}=7.27 \times 10^{-7}\ m$$

A 1.0 g particle is shot from a gun with velocity of 100 m/sec. Calculate its de Broglie wavelength.

We know,

$$\lambda = \dfrac {h}{mu}$$

(Given $$m=1.0g = 10^{-3} kg, \quad u=100 $$ m/sec)

$$\lambda = \dfrac {6.626 \times 10^{-34} }{1\times 10^{-3} \times 100} =6.626 \times 10^{-33}$$m

Calculate the mass of a photon of sodium light having wavelength 5894 $$A^{\circ}$$ and velocity $$=3\times 10^8 ms^{-1} $$, $$h= 6.6 \times 10^{-34}\ kg m^2 s^{-1}$$

Wavelength of photon, $$\lambda =5894 A^{\circ}=5894\times 10^{-10}\ m$$

Velocity of light, $$c= 3 \times 10^8 m s^{-1}$$

Mass of photon $$=\dfrac {h}{c\lambda}= \dfrac {6.6 \times 10^{-34}}{3 \times 10^8 \times 5894\times 10^{-10}}$$

$$=3.73 \times 10^{-36}\ kg$$

The mass of an electron is $$9.1 \times 10^{-31}$$ kg. If its K.E is $$3.0 \times 10^{-25}$$ J, calculate its wavelength in angstrom.

$$K.E= \dfrac {1}{2} mu^2$$ .... (1)

$$\therefore u= \dfrac {h}{m\lambda}$$ .... (2)

By eq (1) and (2),

$$K.E=\dfrac {1}{2} m \dfrac {h^2}{m^2 \lambda ^2}$$ or $$\lambda=\sqrt { \dfrac { { { h^{ 2 } } } }{ { 2m\times K.E } } } $$

$$\lambda=\sqrt { \dfrac {6.626 \times 10^{-34} \times 6.626\times 10^{-34}}{2\times 9.1 \times 10^{-31}\times 3.0 \times 10^{-25} } }$$

$$\lambda=8967\times 10^{-10}m =8967 A^{\circ}$$

Two particle A and B are in motion. If the wavelength associated with particle A is $$ 5\times 10^{-8}$$ m, calculate the wavelength associated with particle B if its momentum if half of A.

$$\lambda_A= \dfrac {h}{p_A}$$ and $$\lambda_B= \dfrac {h}{p_B}$$

$$ \dfrac {\lambda_A}{\lambda_B}=\dfrac {p_B}{p_A}$$ (Given: $${p_B}=\frac{1}{2} {p_A}$$)

$$ \dfrac {5 \times 10^{-8}}{\lambda_B}=\dfrac {1}{2}$$

$$\lambda_B=1\times 10^{-7} \ m$$

Derive Einstein's photoelectric equation $$\dfrac{1}{2}mv^2=hv-hv_0.$$

Einstein explained photoelectric effect on the basis of quantum theory. The main points are
1. Light is propagated in the form of bundles of energy. Each bundle of energy is called a quantum or photon and has energy hv where h=Planck's constant and v=frequency of light.
2 The photoelectric effect is due to collision of a photon of incident light and a bound electron of the metallic cathode.
3. When a photon of incident light falls on the metallic surface, it is completely absorbed, Before being absorbed it penetrates through a distance of nearly $$10^8m(or 100A^\circ).$$ The absorbed photon transfers its whole energy to a single electron. The energy of photon goes in two parts: a part of energy is used in releasinig the electron from the metal surface(i.e., in overcoming work function) and the remaining part appears in the form of kinetic energy of the same electron
If v be the frequency of incident light, the energy of photon =hv. If W be the work function of metal and $$E_k$$ the maximum kinetic energy of photoelectron, then according to Einstein's explanation.
$$hv=W+E_K or E_K=hv-W$$
This is called Einstein's photoelectric equation.
If $$v_0$$ be the threshold frequency, then if frequency of incident light is less than $$v_0$$ no electron will be emitted and if the frequency of incident light be $$v_0$$ then $$E_K=0;$$so from equation(i)
$$0=hv_0-W$$ or $$W=hv_0$$
If $$\lambda_0$$ be the threshold wavelength, then $$v_0=\dfrac{c}{\lambda_0},$$
where c is the speed of light in vacuum
$$\therefore$$Work function $$W=hv_0=\dfrac{hc}{\lambda_0}...(ii)$$
Substituting this value in equation (i), we get
$$E_K-hV=hV_0 \Rightarrow \dfrac{1}{2}mv^2=hv-hv_0...(iii)$$
This is another form of Einstein's photoelectric equation.
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Calculate the wavelength of the electron if it moving with a velocity of $$2.05\times 10^7\ ms^{-1}$$.

According to de Broglie's equation,
$$\lambda=h/mv$$
Where, $$\lambda=$$ wavelength of moving particle
$$m=$$ mass of particle $$(9.109\times 10^{-31}\ Kg)$$
$$v=$$ velocity of particle $$(2.05\times 10^{7}\ ms^{-1})$$
$$h=$$ Planck's constant $$(6.626\times 10^{-34})Js$$

On substituting the values in the expression of $$\lambda$$:
$$\lambda=\cfrac{(6626\times 10^{-34}\ Js)}{(9.109\times 10^{-31})(2.05\times 10^{7}\ ms^{-1})}$$ $$=3.54\times 10^{-11}\ m$$

Hence, the wavelength of the electron is $$3.54\times 10^{-11}\ m$$.

A bulb of $$25\ W$$ emits yellow color monochromatic light of wavelength $$0.57\ \mu m$$. Calculate the number of photons emitted per second

Power of bulb, $$P=25$$ Watt $$=25\ Js^{-1}$$

Energy of a photon is given by $$E=h\nu=hc/\lambda$$.

where, $$c=$$ velocity of light in vaccum $$=3\times 10^{8}\ m/s$$, $$h=$$ Planck's constant $$=6.626\times 10^{-34}\ Js$$, $$\lambda=0.57\times 10^{-6}m$$

On substituting the values, we get,
$$E=(6.626\times 10^{-34})(3\times 10^{8})/0.57\times 10^{-6}$$ $$=34.87\times 10^{-20}\ J$$

Rate of emission of photon per second = $$\cfrac{25}{34.87\times 10^{-20}}=7.169\times 10^{19}$$ photon per second

When radiation of $$6800\ \mathring A$$ wavelength strikes in the metal surface, the electron with zero velocity emitted. Calculate work function $$(\phi)$$ of the metal.

As the kinetic energy of the electron emitted is zero.

The kinetic energy of emitted electron = $$\mathrm{E_{Photon} - \phi}$$

$$\implies \mathrm{0 = E_{Photon} - \phi}$$

$$\implies \mathrm{\phi = E_{Photon}}$$

Energy of a photon is given by: $$\mathrm{E_{Photon} = \cfrac{hc}{\lambda}}$$

$$\implies\mathrm{E_{Photon} = \cfrac{6.62 \times 10^{-34} \times 3 \times 10^8}{6800 \times 10^{-10}} = 5.2 \times 10^{-19} \ J }$$

So, the work function $$(\phi)$$ of the metal = $$5.2 \times 10^{-19} \mathrm{J}$$

Calculate the energy of a photon of light having
(I) frequency $$3\times 10^{15}Hz$$
(ii) the wavelength of the light as $$(0.50\mathring{A})$$

(i) Energy $$(E)$$ of a photon is given by the expression,
$$E=hv$$
where $$h=$$ Planck's constant $$=6.626\times 10^{-34}Js$$, $$v=$$ frequency of light $$=3\times 10^{15}Hz$$ (Given)

On substituting the values we get,
$$E=(6.626\times 10^{-34})\times (3\times 10^{15})$$
$$E=1.988\times 10^{-18}\ J$$

(ii) Energy $$(E)$$ of a photon having wavelength $$(\lambda)$$ is given by the expression,
$$E=\dfrac{hc}{\lambda}$$
where $$h=$$ Palnck's constant $$=6.626\times 10^{-34}\ Js$$, $$c=$$ velocity of light in vaccum $$=3\times 10^{8}\ m/sec$$, $$\lambda=$$ wavelength $$=0.50\mathring A$$

On substituting values in expression we get,
$$E=6.626\times 10^{-34}\times 3\times 10^{8}/0.50\times 10^{-10}$$
$$E=3.98\times 10^{-15}\ J$$

What is the wavelength associated with an electron moving with a velocity of $$\mathrm{4 \times 10^7 \ ms^{-1}}$$?

The wavelength of an object of mass 'm' moving with a velocity 'v' is given by:

$$\mathrm{\lambda = \cfrac{h}{mv}}$$

where 'h' is the plank's constant

The wavelength associated with an electron moving with a velocity of $$\mathrm{4 \times 10^7 \ ms^{-1}} is given by:

$$\mathrm{\lambda = \cfrac{h}{m_ev_e}}$$

$$\mathrm{m_e = 9.1 \times 10^{-31} \ kg \ \ \ \& \ \ \ v_e = 4\times 10^7 \ ms^{-1} }$$

$$\implies \mathrm{\lambda = \cfrac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 4 \times 10^{7}} \ m = 0.18 \times 10^{-10} \ m = 0.18 \mathring A}$$

The work function of the metal is $$2.13\ eV$$. A photon having a wavelength of $$4\times 10^{-7}\ m$$ strikes the metal. Calculate the kinetic energy of the emission of the photon.

The energy of photon is given by: $$\mathrm{E = \cfrac{hc}{\lambda}}$$

Energy of photon having a wavelength of $$4\times 10^{-7}\ m$$ = $$\cfrac{6.62 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}}$$ = $$\mathrm{4.965 \times 10^{-19} \ J}$$

= $$\mathrm{4.965 \times 10^{-19} \times 6.242 \times 10^{18} = 3.1 \ eV}$$

Kinetic Energy of emission of a photon is given by:

$$\mathrm{K.E. = E_{Photon} - \phi}$$

where $$\phi$$ is the work function of metal

$$\implies \mathrm{K.E. = (3.1-2.13) \ eV = 0.97 \ eV}$$

Making use of the uncertainty principle, evaluate the minimum permitted energy of an electron in a hydrogen atom and its corresponding apparent distance from the nucleus.

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Employing the uncertainty principle, evaluate the indeterminancy of the velocity of an electron in a hydrogen atom if the size of the atom is assumed to be $$l = 0.10\ nm$$. Compare the obtained magnitude with the velocity of an electron in the first Bohr orbit of the given atom.

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An excited electron gets relieved of its excess energy by emitting one or more photons of characteristic frequency. Find the uncertainty in the frequency of the radiation if the average period that elapses between the excitation of the atom and the time of radiation is $$10^{-8}$$ second.

The uncertainty in the energy, $$\Delta E$$ of photon is:

$$\Delta E = \dfrac{h/2\pi}{\Delta t} = \dfrac{6.62 \times 10^{-34}}{2 \times 3.14 \times 10^{-8}} = 1.1 \times 10^{-26} J$$

Uncertainty in the frequency, $$\Delta v = \dfrac{\Delta E}{h} = \dfrac{1.1 \times 10^{-26}}{6.62 \times 10^{-34}} = 1.66 \times 10^7 Hz = 16.6 MHz$$

A proton is confined to a nucleus of radius, $$5 \times 10^{-5} m$$. Calculate the minimum uncertainty in its momentum. Also calculate the minimum kinetic energy the proton should have. The proton mass is $$1.6 \times 10^{-27} kg$$.

Diameter of nucleus $$= 2 \times 5 \times 10^{-15} = 10^{-14} m$$.

For a proton to be confined in such a nucleus, the minimum uncertainty in it's position is $$10^{-14} m$$.
Therefore, uncertainty in its momentum, $$\Delta p$$ is given by,
$$\Delta p = \dfrac{h/2 \pi}{\Delta x} = \dfrac{6.62 \times 10^{-34}}{2 \times 3.14 \times 10^{-14}} = 1.05 \times 10^{-20} kg. m/s$$

If above be the uncertainty in protons momentum, the momentum must be atleast comparable in magnitude.
Therefore, $$p = 1.06 \times 10^{-20} kg.m/s$$

Therefore, minimum kinetic energy of proton is given by,
$$KE_{min} = \dfrac{p^2}{2m} = \dfrac{(1.05 \times 10^{-20})^2}{2 \times 1.67 \times 10^{-27}} = 3.3 \times 10^{-14} J = 0.206 MeV$$

Dual Nature Of Radiation And Matter - Class 12 Engineering Physics - Extra Questions

Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (i) red light, and (ii) blue light?

A photosensitive surface emits photoelectrons when red light falls on it. Will the surface emit photoelectrons when blue light is incident on it? Give reason.

Electrons are emitted from the suface when green light is incident on it, but no electrons are ejected when yellow light is incident on it. Do you expect electrons to be ejected when surface is exposed to (i) Red light and(ii) Blue light?

Calculate the de Broglie wavelength of an electron of mass $$9.11 \times 10^{-31}kg$$ and moving with a velocity of $$1.0 \times 10^6 \,ms^{-1}$$.

Define work function for a given metallic surface.

How does the maximum kinetic energy of electrons emitted vary with the work function of the metal?

How does de Broglie wavelength depend on the momentum of a particle?

State Heisenberg's uncertainty principle.

Calculate the de-Broglie wavelength of the electron orbiting in the $$n=2$$ states of hydrogen atom.

Plot the graph between Anode potential and $$\displaystyle\frac{1}{\lambda_{min}}$$. Also find its slope.

Define the term work function of a metal.

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II andindicate your answer by darkening appropriate bubbles in the 4 4 matrix given in the ORS.

Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained using Einstein's equation.

How can photoelectric effect be used to produce electricity.

Give reasons for the followingA 40 watt bulb and 60 watt bulb though have different intensities cannot change the kinetic energy of photoelectrons produced by them.

$$(i)$$ What is thermionic emission?$$(ii)$$ Name the unit in which the work function of a metal is expressed.

Draw a plot showing the variation of photoelectric current with collector plate potential foe two different frequencies, $$v_1 > v_2$$, of incident radiation having the same intensity. In which case will the stopping potential be higher ? Justify your answer.

The following plot is given between $$KE$$ on y-axis and frequency of radiation $$\left(\nu\right)$$ on x-axis. If slope of this plot is $$'m'$$ find ratio $$\left(\displaystyle\frac{m}{h}\right)$$ where $$h = $$ planck's constant.

There are a large number of free electrons moving randomly in a metal. At room temperature, why do they not come out of the surface of metal?

State two important properties of photon which are used to write Einsteins photoelectric equation. Define (i) stopping potential and (ii) threshold frequency, using Einsteins equation and drawing necessary plot between relevant quantities.

The graph shows variation of stopping potential $$V_0$$ versus frequency of incident radiation $$v$$ for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why?

A monochromatic light source of frequency $$6 \times 10^{14} Hz$$ is emitting energy at the rate $$2 \times 10^{-3} J/s$$. Calculate the number of photons emitted per second by the source.

Write Einstein's photoelectric equations and point out any two characteristic properties of photons on which this equations is based.Briefly explain the three observed features which can be explained by this caution.

Write Einstein's photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation.Write the three salient features observed in photoelectric effect which can be explained using this equation.

What is stopping voltage (or cut voltage)? Plot a graph of variation of photoelectric current with collector plate potential for two incident radiations of same frequency and different intensities.

Derive Einstein's photoelectric equation.

The graph between the stopping potential $$V_0$$ and frequency $$v$$ for two different metal plates P and Q are shown in the figure. Which of the metals has greater threshold wavelength and work function?

State Einstein's photoelectric equation. Explain 'two' characteristics on the basis of this equation.

Write any two types of electron emission.

Write Einstein's equation of photoelectric effect. Give Einsteins explanation of photoelectric effect.

Write down Einstein's photoelectric equation.

What is "work function"?

State Moseley's law. What is its importance?

What are de-Broglie waves? Establish the de-Broglie wavelength equation.

The planck's constant is $$h$$ and frequency of a photon is $$\nu$$ then write the formula for Einstein photo electric equation.

The mass of a moving particle is m and velocity is v, then write the formula for de-Broglie wavelength $$\lambda$$.

Write Einstein's Photo-Electric Equation and explain the terms in it.

Albert Einstein, the great physicist proposed a clear picture to explain photoelectric effect.a) Explain Einstein's photo electric equation.b) Name the quanta of light.

Plot a labelled graph of $$|V_s|$$ where $$V_s$$ is stopping potential versus frequency f of the incident radiation.State how will you use this graph to determine the value of Planck's constant.

Show that de Broglie wavelength of electrons accelerated $$V$$ volts is very nearly given by:$$\lambda \left(in \mathring{A} \right) = {\dfrac{150}{V}}^{{1}/{2}}$$.

Why do electron not leave the metal surface on their own?

How can electron be made to leave the metal surface?(State any two ways).

Write Heisenberg's uncertainty principle.

Explain photoelectric effect on the basis of quantum model and derive the photoelectric equation $$hv=\displaystyle\frac{1}{2}mv^2+hv_0$$, where symbols have their usual meanings.

Explain photoelectric emission on the basis of quantum model of light and derive the equation $$hv=\cfrac{1}{2}m{v}^{2}+h{v}_{0}$$, where symbols used have their usual meanings.

Explain photoelectric emission on the basis of quantum model of light and derive Einstein's photoelectric emission. Construct logic symbol, Boolean expression and truth table for AND gate.

Write Einstein's photoelectric effect and explain lens of photoelectric effect with the help of it.

The de-Broglie wave length of a particle moving with a velocity $$2.25 \times 10^8 ms^{-1}$$ is equal to the wavelength of a photon. The ratio of kinetic energy of the particle to the energy of the photon is [velocity of light= $$3\times 10^8 ms^{-1}$$]

Stopping potential [V volts] is plotted against frequency of light used $$[\nu]$$. Find work function(eV). $$(h= 6.62\times 10^{-34} J\times s)$$.

The momentum of an object is doubled. How does its kinetic energy change?

Work function of sodium is $$2.3 eV$$. Does sodium show photo-electric emission for light wavelength $$6800 \mathring {A}$$?

The equation for a wave travelling in x-direction on a string is : y = 3(sin3.14x-314t),thenFind the maximum velocity of particle of the string .

What should be the ratio of velocities of $$CH_4$$ and $$O_2$$ molecules so that they are associated with de Brogile waves of equal wave lenghs ? According to be Broglie equation = $$\lambda = \dfrac{h}{mv}$$

Mention any two types of electron emission.

A particle of mass M at rest decays into two particles of masses $$m_1$$ and $$m_2$$ having non zero velocities. The ratio of the de Broglie wavelengths of the particles, $$\lambda_1/\lambda_2$$ is?

Which colored light bulb - red, orange, yellow, green, or blue- emits photons with (a) the least energy and (b) the greatest energy? Account for your answers.

In the arrangement shown in figure , y=1.0mm, d=0.24mm and D= 1.2m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.

Show that de-Broglie wavelength of electron accelerated through V volt is nearly given by:$$\lambda=[\frac{1550}{V}]^{1/2}$$ in $$A^{\circ}$$

An electron in hydrogen like atom, is in an excited state. It has a total energy of -3.4 eV. Calculate.(a)The kinetic energy of electron(b) The de broglie wavelength of electron$$(h=6.6 \times 10^{-34}, m_e= 9.108 \times 10^{-31}kg)$$

A nonrelativistic particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $$1.813 \times 10^{-4}$$. By calculating its mass, identify the particle.

Write the wave function $$\psi(x)$$ displayed in Eq. $$38-27$$ in the form $$\psi(x)=a+i b,$$ where $$a$$ and $$b$$ are real quantities. (Assume(b) Write the time-dependent wave function $$\Psi(x, t)$$ that $$\psi_{0}$$ is real. that corresponds to $$\psi(x)$$ written in this form.

The uncertainty in the position of an electron along an $$x$$ axis is given as $$50 \mathrm{pm},$$ which is about equal to the radius of a hydrogen atom. What is the least uncertainty in any simultaneous measurement of the momentum component $$p_{x}$$ of this electron?

The frequency v of incident radiation is greater that threshold frequency $$v_0$$ in a photocell. How will the stopping potential vary if frequency (v) is increased, keeping other factors constant?

Two metals A and B have work functions 4eV and 10eV respectively. Which metal has the higher threshold wavelength?

Write the basic features of photon picture of electromagnetic radiation on which Einstein's photoelectric equation is based.

If the intensity of radiation in a photocell is increased how does the stopping potential vary?

Two lines A and B shown in the graph represent the de Broglie wavelength$$(\lambda)$$ as a function of $$\dfrac{1}{\sqrt{V}}$$(V is the accelerating potential) for two particles having the same charge. Which of the two represents the particle of smaller mass?

Chlorophyll present in green leaves of plants absorbs light at $$4.620\times 10^{14}Hz$$. Calculate the wavelength of radiation in nanometer, Which part of the electromagneitc spectrum does it belong to?

Wavelength of different radiations are given below:$$\lambda=(A)300\ nm$$$$\lambda (B)=300\mu\ m$$$$\lambda (C)=3\ nm$$$$\lambda (D)=30\overset { o }{ A } $$Arrange these radiations in the increasing order of their energies.

Explain the working of photo emissive cell. Write any two applications of photoelectric cells.

A hypothetical electromagnetic wave is shown in Fig $$2.2$$. Find out the wavelength of the radiation.

Write Heisenberg Uncertainty principle in relation to position and momentum of particle.

The uncertainty in the X-component of momentum of a moving electrons is $$ 13.18 \times 10^{-30}$$ kg-m/s. Find out the uncertaintities in the X-component of position and velocity.

What explanation was given by Einstein to explain photoelectric effect ? What do you understand by threshold frequency ?

The time period of electromagnetic pulse is $$0.30\,ms$$. Find out the uncertaintity in the energy of photon.

A proton with KE $$6MeV$$ is travelling in a particle accelerator of circular orbit $$0.75m$$. What fraction of energy does it radiate per sec?

The electron energy in hydrogen atom is given by $$E_n$$ = $$\dfrac{-21.7*10^{-19}}{n^2}$$ergs. Calculate the energy required to remove an $$e^-$$ completely from n = 2 orbit. What is the largest wavelength in cm of light that can be used to cause this transition?

An electron moves in an electric field with a kinetic energy of $$2.5 eV$$. What is the associated de Broglie wavelength?

Find the de-Broglie wave length of neutron at $$ {127^\circ }C$$.

Electrons are emitted from the suface when green light is incident on it, but no electrons are ejected when yellow light is incident on it. Do you expect electrons to be ejected when surface is exposed to
(i) Red light and
(ii) Blue light?

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and
indicate your answer by darkening appropriate bubbles in the 4 4 matrix given in the ORS.

Give reasons for the following
A 40 watt bulb and 60 watt bulb though have different intensities cannot change the kinetic energy of photoelectrons produced by them.

$$(i)$$ What is thermionic emission?
$$(ii)$$ Name the unit in which the work function of a metal is expressed.

Write Einstein's photoelectric equations and point out any two characteristic properties of photons on which this equations is based.
Briefly explain the three observed features which can be explained by this caution.

Write Einstein's photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation.
Write the three salient features observed in photoelectric effect which can be explained using this equation.

Albert Einstein, the great physicist proposed a clear picture to explain photoelectric effect.
a) Explain Einstein's photo electric equation.
b) Name the quanta of light.

Plot a labelled graph of $$|V_s|$$ where $$V_s$$ is stopping potential versus frequency f of the incident radiation.
State how will you use this graph to determine the value of Planck's constant.

Show that de Broglie wavelength of electrons accelerated $$V$$ volts is very nearly given by:
$$\lambda \left(in \mathring{A} \right) = {\dfrac{150}{V}}^{{1}/{2}}$$.

The equation for a wave travelling in x-direction on a string is : y = 3(sin3.14x-314t),then
Find the maximum velocity of particle of the string .

What should be the ratio of velocities of $$CH_4$$ and $$O_2$$ molecules so that they are associated with de Brogile waves of equal wave lenghs ?
According to be Broglie equation = $$\lambda = \dfrac{h}{mv}$$

Show that de-Broglie wavelength of electron accelerated through V volt is nearly given by:
$$\lambda=[\frac{1550}{V}]^{1/2}$$ in $$A^{\circ}$$

An electron in hydrogen like atom, is in an excited state. It has a total energy of -3.4 eV. Calculate.
(a)The kinetic energy of electron
(b) The de broglie wavelength of electron
$$(h=6.6 \times 10^{-34}, m_e= 9.108 \times 10^{-31}kg)$$

Wavelength of different radiations are given below:
$$\lambda=(A)300\ nm$$
$$\lambda (B)=300\mu\ m$$
$$\lambda (C)=3\ nm$$
$$\lambda (D)=30\overset { o }{ A } $$
Arrange these radiations in the increasing order of their energies.

Draw a graph showing the variation of stopping potential $$(V_{0})$$ with the $$3$$ frequency of incident radiation $$(v)$$ in relation to the electric effect.
How can the value of Plancks constant be determined from this graph?

Calculate the energy of a photon of light having
(I) frequency $$3\times 10^{15}Hz$$
(ii) the wavelength of the light as $$(0.50\mathring{A})$$