depends on the source of photon

Must be equal to speed of light

May be less than speed of light

May be greater than speed of light

Must be less than speed of light

equal linear momenta have equal wavelengths

equal wavelengths have equal linear momenta

equal energies have equal linear momenta

equal frequencies have equal linear momenta

MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 11

de-Broglie equation is

0%
$n\lambda =2d\quad sin\theta$
0%
$E=hv$
0%
$E={ mc }^{ 2 }$
0%
$\lambda =\dfrac { h }{ mv }$

Explanation

de-Broglie equation is an expression for wavelength and momentum of a particle.

$\lambda=\cfrac{h}{mv}$
Option D is correct.

A proton has kinetic energy

$E=100keV$ which is equal to that of a photon. The wavelength of photon is

${\lambda}_{2}$ and that of proton is

${\lambda}_{1}$ . The ratio of

${\lambda}_{1}/{\lambda}_{2}$ is proportional to

0%
${E}^{2}$
0%
${E}^{1/2}$
0%
${E}^{-1}$
0%
${E}^{-1/2}$

Explanation

For proton and photon wavelength is given as

$\rightarrow$

${ \lambda }_{ proton }=\dfrac { h }{ \sqrt { 2mE } }$

${ \lambda }_{ photon }=\dfrac { { h }_{ c } }{ E }$

$\therefore$ the ratio of

$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } \Rightarrow { E }^{ 1/2 }$

The work function of metal is 1 eV. Light of wavelength 3000 A is incident on this metal surface. The velocity of emitted photo electrons will be

0%
10 m/s
0%
1 x $10^3$ m/s
0%
1 x $10^4$ m/s
0%
1 x $10^6$ m/s

Explanation

We know that,

$E=hv=\dfrac { { h }_{ c } }{ \lambda } =\dfrac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 3\times { 10 }^{ -3 } }$

$E=6.6\times { 10 }^{ -19 }J$

we know that,

$E=\phi +KE$

$6.6\times { 10 }^{ -19 }-1.6\times { 10 }^{ -19 }=\dfrac { 1 }{ 2 } { mV }^{ 2 }\left[ 1eV-16\times { 10 }^{ -19 } \right]$

${ V }^{ 2 }=\dfrac { 5\times { 10 }^{ -19 }\times 2.\quad 1.1\times { 10 }^{ 12 } }{ 9.1\times { 10 }^{ -31 } }$

$\boxed { V=1\times { 10 }^{ 6 }m/s }$

In photoelectric effect, the momentum of incident photon of energy

$3 \times 10^{-19} J$ is :

0%
$9 \times 10^{11} kgms^{-1}$
0%
zero
0%
$10^{-27} kgms^{-1}$
0%
$3 \times 10^{-11} kgms^{-1}$

Explanation

Momentum of photon

$P = \dfrac{E}{c}$

So,

$P = \dfrac{3\times 10^{-19}}{3\times 10^8} = 10^{-27} \ kg \ m/s$

Among the following for which one mathemetical expression

$\lambda =\dfrac { h }{ p }$ stands

0%
De Broglie equation
0%
Einstein
0%
Uncertainty equation
0%
Bohr equation

A photon of energy

$1.12 MeV$ splits into electron - positron pair. The velocity of the electron is (in

$ms^{-1}$ )

0%
$3 \times 10^8$
0%
$4.2 \times 10^8$
0%
$1.32 \times 10^8$
0%
$1.8 \times 10^8$

Two particles of de-broglie wavelength

$\lambda_x$ and

$\lambda_y$ are moving in opposite directions. find dBroglie wavelength after perfectly inelastic collision:

0%
$\dfrac{\lambda_x \lambda_y}{\lambda_x - \lambda_y}$
0%
$\dfrac{2\lambda_x \lambda_y}{\lambda_x - \lambda_y}$
0%
$\dfrac{\lambda_x \lambda^2_y}{\lambda_x - \lambda_y}$
0%
$\lambda_y - \lambda_x$

Explanation

$\dfrac{h}{\lambda_x} - \dfrac{h}{\lambda_y} = \dfrac{h}{\lambda}$

$\dfrac{1}{\lambda} = \dfrac{\lambda_y - \lambda_x}{\lambda_x \lambda_y}$

$\dfrac{\lambda_x \lambda_y}{\lambda_y - \lambda_x}$ or

$\dfrac{\lambda_x \lambda_y}{\lambda_x - \lambda_y}$

Wavelength associated with an electron having kinetic energy

$3 \times 10^{-25} J$ is

$x \times 10^{-7} \ m$ . What is the value of

$x$ ?

0%
9
0%
4
0%
5
0%
6

Explanation

$\begin{array}{l} \lambda =\dfrac { h }{ { \sqrt { 2m\, K.E. } } } =x\times { 10^{ -7 } } \\ \Rightarrow x=\dfrac { { 6.63\times { { 10 }^{ -34 } }\times { { 10 }^{ 7 } } } }{ { \sqrt { 2\times 9\times { { 10 }^{ -31 } }\times 3\times { { 10 }^{ -25 } } } } } \\ =\dfrac { { 6.63\times { { 10 }^{ -27 } } } }{ { \sqrt { 54 } \times { { 10 }^{ -28 } } } } \\ =9.02. \\ Ans.\, (A) \end{array}$

Two particle of mass m and 2m moving in opposite direction collide head on. They have same de-Broglie wavelength before collision after the collision.

0%
If e = 1, de-broglie wavelength of m is greater than that of 2 m
0%
If e = 1, de-Broglie wavelength of 2 m is greater than that of m
0%
The de-Broglie wavelength of m increases if e = 1
0%
The de-Broglie wavelength of 2m remains same if e = 1

A photon possesses:

0%
positive charge.
0%
negative charge.
0%
no charge
0%
depends on the source of photon

Explanation

$\begin{array}{l}\text { A photon is massless, that no } \\\text { electric change and is a stable }\\\text { particle }\end{array}$

Hence option C is correct

A beam of light incident on a surface has photons each of energy

$1 mJ$ and intensity

$25 \, w/cm^2$ . Find number of photons incident per second if surface area is

$25 \, cm^2$ .

0%
$6.25 \times 10^5 \, s^{-1}$
0%
$8.25 \times 10^5 \, s^{-1}$
0%
$6.25 \times 10^4 \, s^{-1}$
0%
$5.25 \times 10^5 \, s^{-1}$

Explanation

$I\times A ={n}\times{Energy \ of \ each \ photon}$

$n=\dfrac{IA}{energy \ of \ each \ photon}$

$\cfrac{25\times 25}{1}\times10^{-3}$

$=6.25\times10^{-5}JS^{-1}$

The ratio of wavelengths of photons emitted when hydrogen atom de-excites from third excited state to second excited state an then de-excites form seconds excited state to first excited state is

0%
$\dfrac {7}{20}$
0%
$\dfrac {20}{7}$
0%
$5$
0%
$20$

Explanation

$\dfrac {1}{\lambda_{1}} = R\left (\dfrac {1}{3^{2}} - \dfrac {1}{4^{2}}\right ) = \dfrac {7R}{144} ..... (1)$

$\dfrac {1}{\lambda_{2}} = R\left (\dfrac {1}{2^{2}} - \dfrac {1}{3^{2}}\right ) = \dfrac {5R}{136} ..... (2)$

$(2)/(1) \dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {5R}{36}\times \dfrac {144}{7R} = \dfrac {20}{7}$ .

If a proton and electron have same de Broglie wavelength, then

0%
Kinetic energy of electron $<$ kinetic energy of proton
0%
Kinetic energy of electron $=$ kinetic energy of proton
0%
Momentum of electron $=$ momentum of proton
0%
Momentum of electron $<$ momentum of proton

Explanation

$\begin{array}{l}\text { If proton and electron have same de-broglie } \\\text { wavelength then momentum of electron is } \\\text { equals to momentum of proton }\end{array}$

Mark the correct statement :

0%
electrons are emitted from metal surface when light of sufficient wavelength falls on it.
0%
the kinetic energy of photo electrons is more for light of longer wavelength in comparison to that due to shorter wavelength.
0%
both of the above
0%
none of the above

The function of photoclecrtic cell is

0%
to convert electrical energy into light energy.
0%
to convert light energy into electrical energy
0%
to convert mechanical energy into electrical energy
0%
to convert DC into AC.

Explanation

$\begin{array}{l}\text { A solar cell or photoelectric cell, is an } \\\text { electrical device that converts the energy } \\\text { of light directly in to electricity by } \\\text { photovoltanic effect }\end{array}$

$\text { light energy } \longrightarrow \text { electrical energy }$

The maximum velocity of the photoelectron emitted by the metal surface is

$v$ . Charge and mass of the photoelectron is denoted by

$e$ and

$m$ respectively. The stopping potential in volt is?

0%
$\dfrac{v^2}{2\left(\dfrac{m}{e}\right)}$
0%
$\dfrac{v^2}{2\left(\dfrac{e}{m}\right)}$
0%
$\dfrac{v^2}{\left(\dfrac{e}{m}\right)}$
0%
$\dfrac{v^2}{\left(\dfrac{m}{e}\right)}$

Explanation

We have,

$\frac{1}{2}mv^{2}_{max} = eV$

V is the stopping potential

$\therefore V= \dfrac{1}{2} \dfrac{v^{2}}{\dfrac{e}{m}}$

The de-Broglie wavelength of electron in

$3^{rd}$ orbit of

$He^{+1}$ ion is approximately.

0%
$2A^0$
0%
$3A^0$
0%
$4A^0$
0%
$5A^0$

Explanation

We know that,

$2\pi r=n\lambda$

$\lambda=\dfrac{2\pi r}{n}$

$\lambda=2\pi\times(0.529\ A^o)\times\dfrac{n^2}{z_{n}}$

$\lambda=2\pi\times(0.529\ A^o)\times\dfrac{n^2}{z}$

$\lambda=2\pi\times(0.529\ A^o)\dfrac{3}{2}$

$\lambda=3\pi\times(0.529\ A^o)≈5\ A^o$

So, Te de broglie wavelength is

$5\ A^{o}$

Hence, D is correct option.

A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths

$'\lambda_x'$ and

$'\lambda_y'$ respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is?

0%
$\lambda_x+\lambda_y$
0%
$\dfrac{\lambda_x\lambda_y}{\lambda_x+\lambda_y}$
0%
$\dfrac{\lambda_x\lambda_y}{|\lambda_x-\lambda_y|}$
0%
$\lambda_x-\lambda_y$

Explanation

By momentum conservation

$P_x-P_y=P_p$

$\dfrac{h}{\lambda_x}-\dfrac{h}{\lambda_y}=\dfrac{h}{\lambda_p}$

$\lambda_p=\dfrac{\lambda_x\lambda_y}{|\lambda_y-\lambda_x|}$ .
1247263_1614068_ans_c81590a0fbe944879caf010d11572e88.png

When photons of energy

$h\nu$ fall on metal plate of work function

$W_o$ , photoelectrons of maximum kinetic energy

$K$ are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be?

0%
$K+W_o$
0%
$K+h\nu$
0%
$K$
0%
$2K$

Explanation

In photoelectric effect,

Total energy = K.E of electrons + Work function

$h\nu = K + W_{0}$

$\therefore\ W_{0} = h\nu - K$

If frequency is doubled, Let max K.E be

$K'$

$\therefore\ h \times 2\nu = K' + W_{0}$

$\therefore\ 2 h\nu = K' +(h\nu - K)$

$\therefore\ K' = hv + K$

At one time the meter was defined as

$1650763.73$ wavelength of the orange light emitted by a light source containing

$Kr^{86}$ atoms. What is the corresponding photon energy of this radiation?

0%
$3.28\times 10^{-19}\ J/quanta$
0%
$1.204\times 10^{-31}\ J/quanta$
0%
$1.09\times 10^{-27}\ J/quanta$
0%
$2.048\ J/quanta$

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius

$4.65 \mathring{A}$ ). The de-Broglie wavelength of this electron is:

0%
$12.9 \mathring{A}$
0%
$3.5 \mathring{A}$
0%
$9.7 \mathring{A}$
0%
$6.6 \mathring{A}$

Explanation

$2\pi r_n = n\lambda_n$

$\lambda_3 = \dfrac{2\pi(4.65 \times 10^{-10})}{3}$

$\lambda_3 = 9.7 {A}^\circ$

in a photoelectric effect experiment the threshold wavelength of the light is

$380nm$ . If the wavelength of incident light is

$260nm$ , the maximum kinetic energy electrons will be:
Given

$E(in \,eV)=\frac{1237}{\lambda(in\,nm)}$

0%
$1.5eV$
0%
$4.5eV$
0%
$15.1eV$
0%
$3.0eV$

Explanation

$K_{max}=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0}$

$\Rightarrow K_{max}=hc(\dfrac{\lambda_0-\lambda}{\lambda \lambda_0})$

$\Rightarrow K_{max}=(1237)(\dfrac{380-260}{380\times260})$

$=1.5eV$

For the same speed, de Broglie wavelength.

0%
Of electron is larger than proton
0%
Of proton is larger than $\alpha-particle$
0%
Of electron is larger than $\alpha-particle$
0%
All of the above

Which of the following statements is false?

0%
Material wave (de Broglie wave) can travel in vacuum
0%
Electromagnetic wave can travel through vacuum
0%
The velocity of photon is not the same whether light passes through any medium
0%
Wavelength of de Broglie wave depends upon velocity

The energy of a photon of frequency

$f$ is

0%
$hf$
0%
$\dfrac {h}{f}$
0%
$h^{2}f$
0%
$h/f^{2}$

The number of ejected photoelectrons increases with increase.

0%
In frequency of light
0%
In wavelength of light
0%
In intensity of light
0%
None of these

Two particles A and B of same mass have their total energies

$E_A$ and

$E_B$ in the ratio

$E_A:E_B=1:2$ . Their potential energies

$U_A$ and

$U_B$ are in the ratio

$U_A:U_B=1:2$ . If

$\lambda_A$ and

$\lambda_B$ are their de-Broglie wavelengths, then

$\lambda_A:\lambda_B$ is

0%
$1:2$
0%
$2:1$
0%
$1:\sqrt{2}$
0%
$\sqrt{2}:1$
0%
$1:1$

Explanation

$\dfrac{E_A}{E_B}=\dfrac{1}{2}$ ,

$\dfrac{U_A}{U_B}=\dfrac{1}{2}$
So

$E_A=x,E_B=2x$
And

$U_A=y,U_B=2y$

$\because E_A=U_A+K_A$
And

$E_B=U_B+K_B$
here

$K_A$ and

$K_B$ are kinetic energy of particles A and B So

$K_A=E_A-U_A=(x-y)$ .........(i)

$K_B=E_B-U_B=2(x-y)$ .........(i)

$\because$ de-Broglie wavelength,

$\lambda=\dfrac{h}{\sqrt{2mk}}$

So,

$\lambda_A=\dfrac{h}{\sqrt{2mK_A}}$ =

$\lambda_B=\dfrac{h}{\sqrt{2mK_B}}$

$\therefore \dfrac{\lambda_A}{\lambda_B}=\sqrt{\dfrac{K_B}{K_A}}$ ...........(iii)
From Equation (i), (ii) and (iii),

$\dfrac{\lambda_A}{\lambda_B}=\sqrt{\dfrac{2(x-y)}{(x-y)}}=\dfrac{\sqrt{2}}{1}$

What wavelength must electromagnetic radiation have if a photon in the beam has the same momentum as an electron moving with a speed

$1.1\times 10^5$ m/s(Planck's constant

$=6.6\times 10^{-34}$ J-s, rest mass of electron

$=9\times 10^{-31}$ kg?

0%
$\dfrac{2}{3}$ nm
0%
$\dfrac{20}{3}$ nm
0%
$\dfrac{4}{3}$ nm
0%
$\dfrac{40}{3}$ nm
0%
$\dfrac{3}{20}$ nm

Explanation

Given,

$m_e=$ mass of electron

$=9\times 10^{-31}kg$

$v_e=1.1\times 10^{5}m/s$

Momentum,

$p=m_ev_e$

$p=9\times 10^{-31}\times 1.1\times 10^5$

$=9.9\times 10^{-26}kg\ m/s$

From de-Broglie waves

$\lambda =\dfrac{h}{p}=\dfrac{6.6\times 10^{-34}}{9.9\times 10^{-26}}m$

$=\dfrac{2}{3}\times 10^{-8}m$

$=\dfrac{20}{3}\times 10^{-9}m=\dfrac{20}{3}nm$

It takes

$4.6eV$ to remove one of the least tightly bound electrons from a metal surface. When monochromatic photons strike the metal surface, electrons having kinetic energy from zero to

$2.2eV$ are ejected. What is the energy of the incident photons?

0%
$2.4eV$
0%
$2.2eV$
0%
$6.8eV$
0%
$4.6eV$
0%
$5.8eV$

Explanation

According to question, the work function of the metal surface is

$W=4.6eV$ and maximum K.E of ejected electron

$=2.2eV$

Let the energy of incident photon is E.

i.e.,

$E=E+K_e$

$K_e=$ Kinetic energy of ejected electrons

$E=4.6eV+2.2eV$

$=6.8eV$ .

The speed of photon.

0%
May be less than speed of light
0%
May be greater than speed of light
0%
Must be equal to speed of light
0%
Must be less than speed of light

If the de Broglie wavelength of a proton is

$10^{-13} m$ , the electric potential through which it must have been accelerated is

0%
$4.07\times 10^{4}V$
0%
$8.2\times 10^{4}V$
0%
$8.2\times 10^{3}V$
0%
$4.07\times 10^{5}V$

What is the de-Broglie wavelength of the

$\alpha$ - particle accelerated through a potential difference

$V$

0%
$\dfrac{0.287}{\sqrt V}\ A^o$
0%
$\dfrac{12.27}{\sqrt V}\ A^o$
0%
$\dfrac{0.101}{\sqrt V}\ A^o$
0%
$\dfrac{0.202}{\sqrt V}\ A^o$

Explanation

As we know,

$\lambda =\dfrac {h}{\sqrt{2mE}}=\dfrac {h}{\sqrt{2m_{\alpha}Q_{\alpha}V}}$
On putting

$Q_{\alpha}=2\times 1.6\times 10^{-19}\ C$

$m_{\alpha }=4m_p=4\times 1.67 \times 10^{-27}\ kg\Rightarrow \lambda =\dfrac {0.101}{\sqrt V}\ A^o$

An electron of mass

$m$ and magnitude of charge

$|e|$ initially at rest gets accelerated by a constant electric field

$E$ . The rate of change of de-Broglie wavelength of this electron at time

$t$ ignoring relativistic effects is:

0%
$\dfrac {-h}{|e| Et^2}$
0%
$-\dfrac {h}{|e| Et}$
0%
$-\dfrac {h}{|e| Et \sqrt t}$
0%
$\dfrac {|e| Et}{h}$

Explanation

$P=\dfrac{h}{\lambda_D}$

$\lambda_D=\dfrac{h}{p}=\dfrac{h}{mv}$

$\because \dfrac{v}{t}=a\Rightarrow v=at$

$v=at$

$\because F=qE$

$ma=qE$

$\boxed{a=\dfrac{qE}{m}=\dfrac{eE}{m}}$

So,

$v=\dfrac{|e|E}{m}t$

$\therefore \lambda_D=\dfrac{h}{m\left(\dfrac{|e|E}{m}\right)t}=\dfrac{h}{|e|Et}$

Rate of change of de-Broglie wavelength is given by differentiating the wavelength w.r.t. time.

$\boxed{\dfrac{d\lambda_D}{dt}=-\dfrac{h}{|e|Et^2}}$

How many photons are emitted by a laser source of

$5\times 10^{-3} W$ operating at

$632.2\ nm$ in

$2$ second?

$(h = 6.63\times 10^{-34} Js)$ .

0%
$3.2\times 10^{16}$
0%
$1.6\times 10^{16}$
0%
$4\times 10^{16}$
0%
None of these

Which one of the following is the correct graph between energy and wavelength for a given photon?

0%
0%
0%
0%
None of these

Photoelectic effect shows.

0%
Wave-like behaviour of light
0%
Particle-like behaviour of light
0%
Both wave-like and particle-like behaviour of light
0%
Neither wave-like nor particle-like behaviour of light

Two photons having

0%
equal wavelengths have equal linear momenta
0%
equal energies have equal linear momenta
0%
equal frequencies have equal linear momenta
0%
equal linear momenta have equal wavelengths

The equation E = pc is valid

0%
for an electron as well as for a photon
0%
for an electron but not for a photon
0%
for a photon but not for an electron
0%
neither for an electron nor for a photon

What is the energy of photon of wavelength

$24800\mathring{A}$ ?

0%
$0.5\ eV$
0%
$0.9\ eV$
0%
$1.1\ eV$
0%
$0.75\ eV$

Two photons of same frequencies moving in same medium have

0%
Same linear momenta and wavelengths
0%
Same linear momenta and same speeds
0%
Same energies and same linear momenta
0%
None of the above

A radiation is incident on a metal surface of work function

$2.3\ eV$ . The wavelength of incident radiation is

$600\ nm$ . If the total energy of incident radiation is

$23\ J$ , then the number of photoelectrons is

0%
Zero
0%
$> 10^{4}$
0%
$= 10^{4}$
0%
None of these

A cylindrical rod of some laser material

$5\times 10^{-2}m$ long and

$10^{-2}m$ in diameter contains

$2\times 10^{25}$ ion per

$m^3$ . If on excitation all the ions are in the upper energy level and de-excite simultaneously emitting photons in the same direction, calculate the maximum energy contained in a pulse of radiation of wavelength

$6.6\times 10^{-7}m$ . If the pulse lasts for

$10^{-7}$ s, calculate the average power of the laser during the pulse

0%
$33.55\ W$
0%
$43.55\ W$
0%
$23.55\ W$
0%
$29.55\ W$

Explanation

From given,

Total no. of ions in the rod
=number of ions per unit volume

$\times$ volume of rod.

$=(2\times10^{25})\times\{3.14\times(0.005)^{2}\times(5\times10^{-2})\}$

The number of photons excited in one direction is equal to the total number of ions because all ions are excited. Now, excited energy

$=(7.85\times10^{19})\times\dfrac{hc}{\lambda}$

$=(7.85\times10^{19})\times\dfrac{6.66\times10^{-34}\times3\times10^{8}}{6.67\times10^{-7}}$

$=23.55\,W$

Five volts of stopping potential is needed for the photoelectrons emitted out of a surface of work function $2.2\,eV$ by the radiation of wavelength

0%
$1719\,\overset{\circ}{A}$
0%
$3444\,\overset{\circ}{A}$
0%
$861\,\overset{\circ}{A}$
0%
$3000\,\overset{\circ}{A}$

Explanation

Einstein's equation for the photoelectric effect is

$eV=\dfrac {hc}{\lambda} - W$

$\dfrac {hc}{\lambda}=5\,eV+2.2\,eV=7.2\,eV$

$7.2=\dfrac{12375}{\lambda(in \overset{\circ}{A})}$

or

$\lambda(in \overset{\circ}{A})=\dfrac{12375}{7.2}=1719\ \overset{\circ}{A}$

The voltage applied to an X-ray tube is

$18 \ kV$ . The maximum mass of photon emitted by the X-ray tube will be

0%
$2 \times 10^{-13} \ kg$
0%
$3.2 \times 10^{-36} \ kg$
0%
$3.2 \times 10^{-32} \ kg$
0%
$9.1 \times 10^{-31} \ kg$

Explanation

Energy of photon is given by

$mc^2$ . Now, the maximum energy of photon is equal to the maximum energy of electrons

$=eV$

Hence,

$mc^2 = eV$

$\implies m= \dfrac{eV}{c^2}$

$= \dfrac{1.6 \times 10^{-19} \times 18 \times 10^3}{(3 \times 10^8)^2} = 3.2 \times 10^{-32} \ kg$

A particle of mass $10^{-31},kg$ is moving with a velocity equal to $10^5 \,ms^{-I}$ . The wavelength of the particle is equal to

0%
$0$
0%
$6.6\times10^{-8}\,m$
0%
$0.66\,m$
0%
$1.5\times10^{-7}\,m$

Explanation

As we know,

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

$\implies\lambda=\dfrac{6.6\times10^{-34}}{10^{-31}\times10^5}$

$\lambda=6.6\times10^{-8}\,m$

Monochromatic light incident on a metal surface emits electrons with kinetic energies
from zero to $2.6\, eV$ . What is the least energy of the
incident photon if the tightly bound electron needs $4.2 \,eV$ to
remove?

0%
$1.6\,eV$
0%
From $1.6\,eV\,to\,6.8\,eV$
0%
$6.8\,eV$
0%
More than $6.8\,eV$

Explanation

According to Einstein's equation,

$E=W_0+KE$

Given,

$W_{0max}=4.2\,eV$

$KE=2.6\,eV$

$E_{min}$ =

$W_{0max}=4.2\,eV+ KE= (4.2+2.6)eV=6.8eV$

The de-Broglie wavelength associated with with the particle of mass

$m$ moving with velocity

$v$ is

0%
$h/mv$
0%
$mv/h$
0%
$mh/v$
0%
$m/vh$

Explanation

The de-Broglie wavelength associated with with the particle of mass

$m$ moving with velocity

$v$ is

$\lambda =\dfrac hp =\dfrac{h}{mv}$

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of

$3\ km/s$ will be

0%
$1\ A^o$
0%
$0.66\ A^o$
0%
$6.6\ A^o$
0%
$66\ A^o$

Explanation

As we know,

$\lambda =\dfrac {h}{mv_{rms}}\Rightarrow \lambda =\dfrac {6.6 \times 10^{-34}}{2\times 1.67 \times 10^{-27}\times 3\times 10^3}=0.66\ A^o$

If we express the energy of a photon in

$keV$ and the wavelength in angstroms, then energy of a photon can be calculated from the relation

0%
$E=12.4\ hv$
0%
$E=1.2\ h/ \lambda$
0%
$E=12.4 / \lambda$
0%
$E=hv$

Explanation

Energy of photon

$E=\dfrac{hc}{\lambda}(joules)=\dfrac{hc}{e \lambda}(eV)$

$\Rightarrow E_(eV)=\dfrac{6.6\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times \lambda (A^o)}=\dfrac{12375}{\lambda (A^o)}$

$\Rightarrow E(keV)=\dfrac{12.37}{\lambda (A^o)} \approx \dfrac{12.4}{\lambda}$

The de-Broglie wavelength of a neutron at

$27^oC$ is

$\lambda$ . What will be its wavelength at

$927^oC$

0%
$\lambda /2$
0%
$\lambda /3$
0%
$\lambda /4$
0%
$\lambda /9$

Explanation

As we know,

$\lambda_{neutron } \propto \dfrac {1}{\sqrt T}\Rightarrow \dfrac {\lambda_1}{\lambda_2}=\sqrt{\dfrac {T_2}{T_1}}$

$\Rightarrow \dfrac {\lambda}{\lambda_2}=\sqrt{\dfrac{(273+927)}{(273+27)}}=\sqrt{\dfrac{1200}{300}}=2\Rightarrow \lambda_2=\dfrac {\lambda}{2}$

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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