MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 5

The energy

$E$ and the momentum

$p$ of a photon is given by

$E = hv$ and

$p = \displaystyle\ \frac{h}{\lambda}$ . The velocity of photon will be

0%
$E/p$
0%
$Ep$
0%
$(E/P)^{2}$
0%
$\sqrt{E/P}$

Explanation

$E=\dfrac{hv}{\lambda}$

also

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

from above two equation

$v=\dfrac{E}{p}$

$10^{-3}W$ of

$5000\overset {o}{A}$ light is directed on a photoelectric cell. If the current in the cell is

$0.16\mu A$ , the percentage of incident photons which produce photoelectrons, is

0%
40%
0%
0.04%
0%
20%
0%
10%

Explanation

Energy of each photon

$E=\dfrac { hc }{ \lambda } =\dfrac { 1.986\times { 10 }^{ -25 } }{ 5000\times { 10 }^{ -10 } } =3.936\times { 10 }^{ -19 }$

Number of Photons per sec:

$N=\dfrac { { 10 }^{ -3 } }{ { 3.936\times { 10 }^{ -19 } } } =2.54\times { 10 }^{ 15 }$

Number of electrons per sec:

$N'=\dfrac { 0.16\times { 10 }^{ -6 } }{ 1.6\times { 10 }^{ -19 } } ={ 10 }^{ 12 }$

$\therefore \quad \dfrac { N' }{ N } =\dfrac { { 10 }^{ 12 } }{ 2.54\times { 10 }^{ 15 } } \approx 4\times { 10 }^{ -4 }$

In percentage

$\dfrac { N' }{ N } \times 100=0.04$ %

If 5% of the energy supplied to a bulb is irradiated as visible light, how many quanta are emitted per second by a 100 W lamp? Assume wavelength of visible light of

$5.6\times 10^{-5} cm$ .

0%
$1.4\times 10^{19}$
0%
$3\times 10^{3}$
0%
$1.4\times 10^{-19}$
0%
$3\times 10^{4}$

Explanation

Let n be the emitted photons per sec.

We know that photon energy ,

$E=h\nu=\dfrac{hc}{\lambda}$

here,

$nE=5$

So,

$n=\dfrac{5\lambda}{hc}=\dfrac{5\times 5.6\times 10^{-7}}{(6.62\times 10^{-34})\times (3\times 10^8)}=1.4\times 10^{19}$

The radius of the second orbit of an electron in hydrogen atom is

$2.116\overset {o}{A}$ . The de Broglie wavelength associated with this electron in this orbit would be

0%
$6.64\overset {o}{A}$
0%
$1.058\overset {o}{A}$
0%
$2.116\overset {o}{A}$
0%
$13.28\overset {o}{A}$

Explanation

From Bohr's postulate,

$mvr=\dfrac{nh}{2\pi}$ ... (1) where

$m=$ mass of electron,

$v=$ orbital velocity of electrons ,

$r=$ radius of the orbit and

$h=$ Planck's constant

de Broglie wavelength,

$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$ ....(2)

From (1) and (2),

$\lambda=\dfrac{2\pi r}{n}=\dfrac{2\times 3.14\times 2.116}{2}=6.64 A^o$

$\lambda_1$ and

$\lambda_2$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom, then

$\lambda_1/\lambda_2$ is equal to :

0%
$2$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$4$

Explanation

The De-Broglie wavelengths can be given by:

$\lambda =\dfrac{h}{mv}..(1)$

Since mass of electron remains same, we get

$\dfrac{\lambda _{1}}{\lambda _{2}}=\dfrac{v_{2}}{v_{1}}..(2)$

Now, by Bohr's postulates, we know that velocity of electron in

$n^{th}$ orbit in inversely proportional to

$n$ .

Hence, Answer=

$1/2$ .

A particle of mass M at rest decays into two masses

$m_1$ and

$m_2$ with non-zero velocities. The ratio

$\lambda_1 / \lambda_2$ of de Broglie wavelengths of particles is

0%
$m_2/m_1$
0%
$m_1/m_2$
0%
$\sqrt {m_1}/\sqrt {m_2}$
0%
1:1

Explanation

By momentum conservation,

$Mv=m_1v_1+m_2v_2 ...(1)$

As M is initially rest so

$v=0$

Now (1) becomes,

$0=m_1v_1+m_2v_2$

$m_1v_1=-m_2v_2$ , it means that the magnitude of momentum of both masses are equal i.e

$p_1=p_2$

From de Broglie hypothesis,

$\lambda_1=h/p_1$ and

$\lambda_2=h/p_2$

So,

$\dfrac{\lambda_1}{\lambda_2}=\dfrac{p_2}{p_1}=1$ (as

$p_1=p_2$ )

A particle of mass m is projected from ground with velocity u making angle

$\theta$ with the vertical. The de Broglie wavelength of the particle at the highest point is

0%
$\infty$
0%
$h/mu sin\theta$
0%
$h/mu cos\theta$
0%
$h/mu$

Explanation

De Broglie wavelength is given by:

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

At the highest point, only the horizontal component of the velocity will be there, which is

$u\sin{\theta}$ .

Thus, wavelength is given by,

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

What is the wavelength of a photon of energy 1eV?

0%
$12.4\times 10^3\overset {o}{A}$
0%
$2.4\times 10^3\overset {o}{A}$
0%
$0.4\times 10^2\overset {o}{A}$
0%
$1000\overset {o}{A}$

Explanation

Energy of photon

$E = 1 \ eV$
Wavelength of photon

$\lambda (in \ A^o) = \dfrac{hc}{E} = \dfrac{12400}{E \ (in \ eV)}$

$\implies \ \lambda = \dfrac{12400}{1} \ A^o =12.4\times 10^3 \ A^o$

How many photons of a radiation of wavelength

$\lambda=5\times 10^{-7}$ m must fall per second on a blackened plate in order to produce a force of

$6.62\times 10^{-5}N$ ?

0%
$3\times 10^{19}$
0%
$5\times 10^{22}$
0%
$2\times 10^{22}$
0%
$1.67\times 10^{18}$

Explanation

Momentum of photon is given by,

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

Momentum of n photons,

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

Force is the change in momentum. Since, the plate is blackened, all the photons are being absorbed. Thus, change in momentum and hence the force is given by

$F=\dfrac{dp}{dt}=n\dfrac{h}{\lambda}$

$n=F\dfrac{\lambda}{h}=\dfrac{6.62\times{10}^{-5}\times 5\times {10}^{-7}}{6.62\times{10}^{-34}}=5\times{10}^{22}$

A particle of mass 3m at rest decays into two particles of masses m and 2m having non-zero velocities. The ratio of the de Broglie wavelengths of the particles

$(\lambda_1/\lambda_2)$ is

0%
1/2
0%
1/4
0%
2
0%
None of these

Explanation

By conservation momentum,

$3mv=mv_1+2mv_2$

At mass 3m is at rest so

$v=0$

Now,

$0=mv_1+2mv_2$ or

$p_1+2p_2=0$ where

$p_1, p_2$ are the momentum of m and 2m.

Thus,

$|p_1|=2|p_2| ...(1)$

So,

$\lambda_1=\dfrac{h}{|p_1|}$ and

$\lambda_2=\dfrac{h}{|p_2|}$

Thus,

$\dfrac{\lambda_1}{\lambda_2}=\dfrac{|p_2|}{|p_1|}=1/2$ using (1)

How many photons are emitted per second by a 5 mW laser source operating at 632.8 nm?

0%
$1.6\times 10^{16}$
0%
$1.6\times 10^{13}$
0%
$1.6\times 10^{10}$
0%
$1.6\times 10^{3}$

Explanation

The energy of a photon is

$E=\dfrac{hc}{\lambda}=\dfrac{(6.62\times 10^{-34})(3\times 10^8)}{632.8\times 10^{-9}}=3.14\times 10^{-19 } J$

Given, the energy emitted per second by laser

$=5 mW=5\times 10^{-3} J$

Thus the number of photons emitted per second

$=\dfrac{5\times 10^{-3}}{3.14\times 10^{-19}}=1.6\times 10^{16}$

The de Broglie wavelength of a thermal neutron at

$927^oC$ is

$\lambda$ . Its wavelength at

$327^oC$ will be

0%
$\lambda /2$
0%
$\lambda /\sqrt 2$
0%
$\lambda \sqrt 2$
0%
$2\lambda$

Explanation

The relation between de Broglie wavelength and temperature is given by,

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

where

$k=$ Boltzmann's constant ,

$m=$ mass,

$T=$ temperature.

Here all are constant except

$T$ so

$\lambda \propto \dfrac{1}{\sqrt T}$

Thus,

$\dfrac{\lambda_2}{\lambda_1}=\sqrt{\dfrac{T_1}{T_2}}=\sqrt{\dfrac{273+927}{273+327}}=\sqrt 2$

$\lambda_2=\lambda \sqrt 2$

The energy of a photon is equal to the kinetic energy of a proton. The energy of photon is

$E$ . Let

$\lambda_1$ be the de Broglie wavelength of the proton and

$\lambda_2$ be the wavelength of the photon. Then,

$\lambda_1/\lambda_2$ is proportional to :

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

Explanation

The energy photon ,

$E=h\nu_2=\dfrac{hc}{\lambda_2}$ or

$\lambda_2=\dfrac{hc}{E}$

As energy of photon

$=$ energy of proton so

$E=\dfrac{p^2}{2m}$ where p be the momentum of proton.

Now,

$\lambda_1=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mE}}$

Thus,

$\dfrac{\lambda_1}{\lambda_2}=\dfrac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}}=\dfrac{1}{c\sqrt{2m}}\sqrt E$

Hence,

$\dfrac{\lambda_1}{\lambda_2}\propto E^{1/2}$

A helium-neon laser has a power output of 1 mW of light of wavelength 632.8 nm. Calculate the energy of each photon in eV.

0%
2.5
0%
1.96
0%
0.53
0%
3.3

Explanation

The energy of a photon is

$E=\dfrac{hc}{\lambda}$ where

$h=$ Planck's constant,

$c=$ velocity of light.

So,

$E=\dfrac{(6.62\times 10^{-34})(3\times 10^8)}{632.8\times 10^{-9}}=3.14\times 10^{-19}J=1.96 eV$ where

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

Two hydrogen atoms are in excited state with electrons residing in

$n=2$ . The first one is moving toward left and emits a photon of energy

$E_1$ toward right. The second one is moving toward right with the same speed and emits a photon of energy

$E_2$ toward left. Taking recoil of nucleus into account, during the emission process

0%
$E_1>E_2$
0%
$E_1< E_2$
0%
$E_1=E_2$
0%
Information insufficient

Explanation

Energy obtained due to electron transition will be same, but nucleus moving in same direction imparts more K.E.
Hence,

${ E }_{ 1 }<{ E }_{ 2 }$

A 100 W point source emits monochromatic light of wavelength

$6000\overset {o}{A}$ . Calculate the total number of photons emitted by the source per second.

0%
$5\times 10^{20}$
0%
$8\times 10^{20}$
0%
$6\times 10^{20}$
0%
$3\times 10^{20}$

Explanation

The energy of the light

$E=\dfrac{hc}{\lambda}=\dfrac{(6.62\times 10^{-34})(3\times 10^8)}{6000\times 10^{-10}}=3.31\times 10^{-19} J$

We know that power is the energy per second. Thus, the energy per second is given

$100 J$

Thus, number of photons per second

$=\dfrac{100}{3.31\times 10^{-19}}=3.02\times 10^{20}\sim 3\times 10^{20}$

An electron and a photon have same wavelength. If

$p$ is the momentum of electron and

$E$ is the energy of photon, the magnitude of

$p/E$ in

$S\space I$ unit is

0%
$3.0\times10^8$
0%
$3.33\times10^{-9}$
0%
$9.1\times10^{-31}$
0%
$6.64\times10^{-34}$

Explanation

From DeBroglie relation

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

and using energy relation

$E = \dfrac{hc}{\lambda}$ or

$\lambda = \dfrac{hc}{E}$

Equating these two

$\dfrac{h}{p} = \dfrac{hc}{E}$

hence ,

$\dfrac{p}{E} = \dfrac{1}{c} = 3.33 * 10^{-9}$

An electron is in an excited state in a hydrogen like atom. It has a total energy of

$-3.4\space eV$ . The kinetic energy of electron is

$'E\ '$ and its de Broglie wavelength is

$'\lambda\ '$

0%
$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-10}\space m$
0%
$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-10}\space m$
0%
$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-11}\space m$
0%
$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-11}\space m$

Explanation

In any state,

$KE = E, and PE = -2E$ , hence the total energy is

$-E.$
Here

$E = 3.4eV.$
Converting

$E$ into

$Joules$ and using the relation.

$E = \dfrac{hc}{\lambda}$

$\lambda = \dfrac{hc}{E} = 6.6 * 10^{-10} m$

The energy of photon of green by Potential DIfference of

$5000A^0$ . is

0%
$3.459 \times 10^{-19} joule$
0%
$3.973 \times 10^{-19} joule$
0%
$4.132 \times 10^{-19} joule$
0%
$8453 \times 10^{-19} joule$

Explanation

$E=hc/\lambda$

$=6.6 \times 10^{-34}\times 3\times 10^8/5000\times 10^{-10}$

$=3.973 \times 10^{-19}\, J$

An electron of mass 'm' and charge 'w' 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%
$-\dfrac{h}{eEt^2}$
0%
$-\dfrac{eht}{E}$
0%
$-\dfrac{mh}{eEt^2}$
0%
$\dfrac{h}{eE}$

Explanation

ginen,

$u=0$

$F = Ee$

$a =\cfrac{F}{m} = \cfrac{Ee}{m}$
Newton's equation of motion

$v=u+at$

$v= \cfrac{Eet}{m}$
Debroglie Wavelength

$\lambda = \cfrac{h}{mv} = \cfrac{h}{Eet}$
Rate of change

$\cfrac{d \lambda}{dt} = -\cfrac{h}{Eet^2}$

We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used the energy required should be

0%
1.5 keV
0%
15 keV
0%
150 keV
0%
1.5 MeV

Explanation

De broglie wavelength is given by-

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

$\implies v=\dfrac{h}{m\lambda}$

$\lambda=10pm=10^{-11}m$

Energy required=

$\dfrac{1}{2}mv^2=\dfrac{1}{e}\dfrac{1}{2}m(\dfrac{h}{m\lambda})^2eV$

$=15keV$

The energy of photon of wavelength

$\lambda$ is

0%
$c \lambda /h$
0%
$h \lambda /c$
0%
$hc/ \lambda$
0%
$\lambda/ hc$

Explanation

Energy of photon =

$\dfrac{hc}{\lambda}$

Hence, a photon with longer wavelenght will carry less energy.

The wavelength of a wave is

$\lambda$ = 6000

$\overset{o}{A}$ , then wave number will be

0%
1.66 $\times$ 10 $^{7}$ m $^{-1}$
0%
1.66 $\times$ 10 $^{6}$ m $^{-1}$
0%
16.6 $\times$ 10 $^{-1}$ m $^{-1}$
0%
166 $\times$ 10 $^{3}$ m $^{-1}$

Explanation

Wave number,

$\displaystyle\bar{v}$ =

$\displaystyle\dfrac{1}{\lambda}$

$= \dfrac{1}{6 \times 10^{-7}}$

$= 1.66 \times 10^{6} m^{-1}$

A proton and

$\alpha$ -particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelength will be

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

Explanation

De Broglie wavelength is given by:

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

Writing momentum as a a function of kinetic energy and mass

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

and Kinetic energy = potential through which it is accelerated times the charge on it

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

$\dfrac{\lambda_{proton}}{\lambda_{alpha}} = \sqrt{\dfrac{q_{alpha}m_{alpha}}{q_{proton}m_{proton}}}$

$q_{alpha} = 4e$

$q_{proton} = e$

$m_{alpha} = 4m_{proton}$

Hence,

$\dfrac{\lambda_{proton}}{\lambda_{alpha}} = \sqrt{\dfrac{8}{1}} = 2\sqrt{2}$

If the kinetic energy of a moving particle is E, then the de-Broglie wavelength is

0%
$\lambda =h \sqrt{2 m E}$
0%
$\lambda =\sqrt{\dfrac{2 m E}{h}}$
0%
$\lambda =\dfrac{h}{\sqrt{2 m E}}$
0%
$\lambda =\dfrac{hE}{\sqrt{2 m E}}$

Explanation

$E=\dfrac{1}{2}mv^2 \,\,or\,\,$

$mv= \sqrt{2mE} \,\, or\,\,$

$\lambda =\dfrac{h}{mv}=\dfrac{h}{\sqrt{2mE}}$

If a photon and an electron have same de-broglie wavelength, then

0%
Both have same kinetic energy
0%
Proton has more K.E. than electron
0%
Electron has more K.E. than proton
0%
Both have same velocity

Explanation

De Broglie wavelength is given by:

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

$p = \sqrt{2Em}$ Where E is kinetic energy.
Given that both the electron and proton have the same De Broglie wavelength.
Then

$p_{e} = p_{p}$

$E_{electron}m_{electron} = E_{proton}m_{proton}$

$\dfrac{E_{electron}}{E_{proton}} = \dfrac{m_{proton}}{m_{electron}}$

and we know that:

$\dfrac{m_{proton}}{m_{electron}} > 1$

Hence

$E_{electron} > E_{proton}$

If the energy of a photon is 10 eV, then its momentum is :

0%
$5.33 \times 10^{-23} kg \,m/s$
0%
$5.33 \times 10^{-25} kg \,m/s$
0%
$5.33 \times 10^{-29} kg \,m/s$
0%
$5.33 \times 10^{-27} kg \,m/s$

Explanation

Momentum of a photon

$= \dfrac{E}{c}= \dfrac{10\times 1.6\times 10^{-19}}{3\times 10^8}$

$= 5.33 \times 10^{-27} kg \,m/s$

It is essential to consider light as a stream of photons to explain

0%
Diffraction of light
0%
Refraction of light
0%
Photoelectric effect
0%
Reflection of light

A 200W sodium street lamp emits yellow light of wavelength

$0.6\mu m$ . Assuming it to be 25% efficient converting electrical energy to light, the number of photons of yellow light it emits per second is :

0%
$62\times 10^{20}$
0%
$3\times 10^{19}$
0%
$1.5\times 10^{20}$
0%
$6\times 10^{18}$

Explanation

The number of photons per sec

$, n=$ amount of power converted to light / energy of photon

Power converted to light

$P_l=200(25/100)=50 W$

Energy of photon,

$E_p=\dfrac{hc}{\lambda}=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{0.6\times 10^{-6}}=33.1\times 10^{-20} J$

Thus,

$n=\dfrac{50}{33.1\times 10^{20}}=1.5\times 10^{20}$

Two hydrogen atoms are in excited state with electrons residing in

$\displaystyle n=2$ . First one is moving towards left and emits a photon of energy

$\displaystyle { E }_{ 1 }$ towards right. Second one is moving towards left with same speed and emits a photon of energy

$\displaystyle { E }_{ 2 }$ towards left. Taking recoil of nucleus into account during emission process, which of the following option is correct?

0%
$\displaystyle { E }_{ 1 }>{ E }_{ 2 }$
0%
$\displaystyle { E }_{ 1 }<{ E }_{ 2 }$
0%
$\displaystyle { E }_{ 1 }={ E }_{ 2 }$
0%
information insufficient

Explanation

From conservation of momentum in both cases,

$(i) mv'-\dfrac{h\nu}{c}=mv$

$(ii) mv''+\dfrac{h\nu'}{c}=mv$

Thus

$mv'-\dfrac{h\nu}{c}=mv''+\dfrac{h\nu'}{c}$

Thus

$v'>v'''$

Also from conservation of energy

$\dfrac{1}{2}mv'^2+h\nu=\dfrac{1}{2}mv''^2+h\nu'$

Since

$v'>v'', \nu<\nu'$

$\implies E_1<E_2$

In photoelectric effect, ______ present in solar energy changes into electric energy.

0%
Only radiant heat
0%
Visible light
0%
Both radiant heat and light
0%
Neither radiant heat nor light

The condition for achieving laser action are
(i) the system must be in a state of population inversion
(ii) the excited state of the system should be in metastable state
(iii) the atom should be in lower energy state
(iv) no conditions required

0%
(i) and (ii)
0%
(ii) and (iii)
0%
(iii) and (iv)
0%
(i), (ii), (iii),(iv)

Explanation

Laser is light amplification by stimulated emission radiation.

For stimulated emission to take place, there should be more no. of atoms in the higher energy state than ground state. This non-equilibrium between the energy states is referred as population inversion. This is important for laser action to take place.

The population inversion occurs only in between the metastable state and lower state. Metastable state is a higher state but with a lifetime of

$10^{-5}$ to

$10 ^{-6} s$ . It is a short lived state but longer lifetime than an ordinary excited state and shorter lifetime than ground state.

LASER action is found in _________ semiconductor.

0%
direct bond gap
0%
indirect bond gap
0%
germanium
0%
silicon

In LASER during the stimulated emission process the photon is __________.

0%
lost
0%
created
0%
absorbed
0%
scattered

If the energy of photons corresponding the wavelength of

$6000\mathring { A }$ is

$3.2\times {10}^{-19}J$ , the photon energy for a wavelength of

$4000\mathring { A }$ will be

0%
$1.11\times {10}^{-19}J$
0%
$2.22\times {10}^{-19}J$
0%
$4.44\times {10}^{-19}J$
0%
$4.80\times {10}^{-19}J$

Consider two particles of different masses. In which of the following situations the heavier of the two particles will have smaller de-Broglie wavelength?

0%
Both have a free fall through the same height
0%
Both moves with the same kinetic energy
0%
Both moves with the same linear momentum
0%
Both move with the same speed

Explanation

$\lambda =\frac { h }{ p }$ , thus, more the momentum the particle smaller is the de-Broglie wavelength. In option A, the speed of the particles is same, thus, the heavier particle will have more momentum. in option B, it can be seen that the heavier particle will have more linear momentum. in option D, they have the same speed , thus, heavier particle will have more linear momentum.

The momentum of a photon of energy

$1MeV$ in kg-m/s, will be equal to :

0%
$0.33\times {10}^{6}$
0%
$7\times {10}^{-24}$
0%
${10}^{-22}$
0%
$5\times {10}^{-22}$

Explanation

Energy of photon is given by

$E=\cfrac{hc}{\lambda}............(i)$

where

$h$ is the Planck's constant,

$c$ the velocity of light and

$\lambda$ its wavelength.

de-Broglie wavelength is given by

$\lambda=\cfrac{h}{p}............(ii)$

where

$p$ is being momentum of photon.

From Eqs.

$(i)$ and

$(ii)$ , we get

$E=\cfrac{hc}{h/p}=pc$

$p=E/c$

Given

$E=1$

$Mev=1\times {10}^{6}\times 1.6\times {10}^{-19}J$ ,

$c=3\times {10}^{8}m/s$

Hence, after putting numerical values we obtain

$p=\cfrac{1\times {10}^{6}\times 1.6\times {10}^{-19}}{3\times {10}^{8}}kg-m/s$

$=5\times {10}^{-22}kg-m/s$

For the Bohr's first orbit of circumference

$2\pi r$ , the de-Broglie wavelength of revolcing electron will be.

0%
$2\pi r$
0%
$\pi r$
0%
$\displaystyle\frac{1}{2\pi r}$
0%
$\displaystyle\frac{1}{4\pi r}$

Explanation

$mvr=\displaystyle \frac{nh}{2\pi}$ , according to Bohr's theory

$\Rightarrow 2\pi r=n\left(\displaystyle\frac{h}{mv}\right)=n\lambda$ for

$n=1, \lambda =2\pi r$

The energy of gamma

$\left( \gamma \right)$ ray photon is

${ E }_{ \gamma }$ and that of an X-ray photon is

${ E }_{ X }$ . If the visible light photon has an energy of

${ E }_{ v }$ , then we can say that:

0%
${ E }_{ X }>{ E }_{ \gamma }>{ E }_{ v }$
0%
${ E }_{ \gamma }>{ E }_{ v }>{ E }_{ X }$
0%
${ E }_{ \gamma }>{ E }_{ X }>{ E }_{ v }$
0%
${ E }_{ X }>{ E }_{ v }>{ E }_{ \gamma }$

Explanation

Energy of a photon

$E = h\nu$

where

$\nu$ is the frequency of the photon and

$h$ is Planck's constant

$\implies$

$E \propto \nu$

$\nu_{\gamma} > \nu_{X} > \nu_v$

$\implies$

$E_{\gamma} > E_{X} >E_v$

The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is :

0%
$\pi r^2$
0%
$2\pi r$
0%
$\pi r$
0%
$\sqrt{\pi r}$

Explanation

Bohr's postulate for angular momentum states,

$mvr=\dfrac{nh}{2\pi}$

For ground state, momentum of electron

$=p=mv=\dfrac{h}{2\pi r}$

Thus de-broglie wavelength for the electron is

$\dfrac{h}{p}=2\pi r$

In photoelectric effect if the intensity of light is doubled, then maximum kinetic energy of photoelectrons will become.

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Double
0%
Half
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Four times
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No change

When the kinetic energy of an electron is increased the wavelength of the associated wave will :

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Increase
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Decrease
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Wavelength does not upon kinetic energy
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None of the above

Explanation

Since the kinetic energy of electron is given as

$E=\dfrac{p^2}{2m}$

where

$p$ is the momentum of the electron

$\implies p=\sqrt{2mE}$

Thus de-broglie wavelength=

$\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mE}}$

Hence as energy of electron increases, the wavelength associated decreases.

Hence correct answer is option B.

The energy of an electron of mass m moving with velocity V and de-Broglie wavelength

$\lambda$ is __________. ('h' is Planck's constant)

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$\displaystyle\frac{h}{2m\lambda}$
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$\displaystyle\frac{h^2}{2m\lambda^2}$
0%
$\displaystyle\frac{h\lambda}{2m}$
0%
$\displaystyle\frac{h}{m\lambda}$

Explanation

Since the electron is not moving in an electric field, its potential energy is zero and its total energy is equal to kinetic energy.

$E=\dfrac { 1 }{ 2 } m{ V }^{ 2 }=\dfrac { 1 }{ 2 } m{ (\dfrac { h }{ m\lambda } ) }^{ 2 }=\dfrac { h^{ 2 } }{ 2m{ \lambda }^{ 2 } }$

The de-Broglie wavelength of an electron moving with a velocity

$\dfrac{c}{2}$ (where c is velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is:

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$1 : 4$
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$1 : 2$
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$1 : 1$
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$2 : 1$

Explanation

De-broglie wavelength of the electron

$=\dfrac{h}{m_ev}=\dfrac{h}{m_e(c/2)}$

Wavelength of photon

$=\lambda$ (say)

Thus,

$\lambda=\dfrac{h}{m_e(c/2)}$

Ration of kinetic energies

$=\dfrac{\dfrac{1}{2}m_e(\dfrac{c}{2})^2}{h\dfrac{c}{\lambda}}=\dfrac{\dfrac{1}{2}m_e(\dfrac{c}{2})^2}{m_e\dfrac{c^2}{2}}=1:4$

$1$ mole of photon, each of frequency

$2500{ s }^{ -1 }$ , would have approximately a total energy of :

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$10 erg$
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$1 J$
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$1 eV$
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$1 MeV$

Explanation

Frequency of one photon

$=2500{ s }^{ -1 }$

$\therefore$ Frequency of

$1$ mole of photon

$=6.023\times { 10 }^{ 23 }\times 2500{ s }^{ -1 }$
Total energy of

$1$ mole of photon,

$E = hv$

$=6.626\times { 10 }^{ -34 }Js\times 6.023\times { 10 }^{ 23 }\times 2500{ s }^{ -1 }$

$=9.97\times { 10 }^{ -7 }J\approx 10erg$

Which of the following expression gives the de-Broglie relationship?

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$\displaystyle p=\frac { h }{ mv }$
0%
$\displaystyle \lambda =\frac { h }{ mv }$
0%
$\displaystyle \lambda =\frac { h }{ mp }$
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$\displaystyle \lambda m=\frac { v }{ p }$

Explanation

The de-Broglie relation is,

$\displaystyle \lambda =\frac { h }{ mv }$
Where,

$\displaystyle \lambda$ = de-Broglie wavelength

$\displaystyle h$ =Planck's constant

$\displaystyle m$ =mass of particle
and

$\displaystyle v$ = velocity of particle

Hard X-rays for the study of fractures in bones should have a minimum wavelength of

$\displaystyle { 10 }^{ -11 }m$ . The accelerating voltage for electrons in X-ray machine should be:

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$<124 \ kV$
0%
$>124 \ kV$
0%
Between $60 \ kV$ and $70\ kV$
0%
$=100\ kV$

Explanation

From conservation of energy the electron kinetic energy equals the maximum photon energy (we neglect the work function

$\displaystyle \phi$ because it is normally so small compared to

$\displaystyle { eV }_{ 0 }$

$\displaystyle { eV }_{ 0 }={ hv }_{ max }$

$\displaystyle { eV }_{ 0 }=\frac { hc }{ { \lambda }_{ min } }$

$\displaystyle { V }_{ 0 }=\frac { hc }{ { e\lambda }_{ min } }$

$h$ (Plank's Constant) =

$6.625\times 10^{-34}J-s$ ,

$c$ (Speed of light) =

$3\times 10^{8}\$

$m/s$ ,

$e$ (Charge of electron) =

$1.6\times10^{-19}C$

$\displaystyle { V }_{ 0 }=\frac { 12400\times { 10 }^{ -10 } }{ { 10 }^{ -11 } }$

$\displaystyle =124\ kV$
Hence, accelerating voltage for electrons in X-ray machine should be less than

$124\$

$kV$ .

The energy of a photons is equal to the kinetic energy of a proton. If

$\lambda_1$ is the de-Broglie wavelength of a proton,

$\lambda_2$ the wavelength associated with the photon, and if the energy of the photon is E, then

$(\lambda_1/\lambda_2)$ is proportional to:

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$E^4$
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$E^{1/2}$
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$E^2$
0%
$E$

Explanation

Given :

$E_{proton} = E_{photon} =E$

de-Broglie wavelength of the proton

$\lambda_1 = \dfrac{h}{p}$ where

$p = \sqrt{2mE}$

$\implies$

$\lambda_1 = \dfrac{h}{\sqrt{2mE}}$ .......(1)

Wavelength associated with photon is

$\lambda_2$

$\therefore$

$E = \dfrac{hc}{\lambda_2}$

$\implies \lambda_2 =\dfrac{hc}{E}$ .........(2)

$\therefore$

$\dfrac{\lambda_1}{\lambda_2} = \dfrac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}} = \dfrac{\sqrt{E}}{c\sqrt{2m}}$

$\implies$

$\dfrac{\lambda_1}{\lambda_2} \propto E^{1/2}$

The energy that should be added to an electron to reduce its de-Broglie wavelength from

$1$ nm to

$0.5$ nm is.

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Four times the initial energy
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Equal to the initial energy
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Twice the initial energy
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Thrice the initial energy

Explanation

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

$\dfrac{\lambda'}{\lambda} = \dfrac{\sqrt E}{E'}$

$\therefore \dfrac{E}{E'} = (\dfrac{\lambda'}{\lambda})^2$

$= (\dfrac{0.5}{1})^2 = \dfrac{1}{4}$

so E' = 4E

so difference is 3E

Calculate the approximates energy of a photon given that wavelength of photon is 2 nm and Planck constant

$h$ is

$6.6\times 10^{-34} Js$ .

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$4\times 10^{-51}J$
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$1\times 10^{-34}J$
0%
$1\times 10^{-16}J$
0%
$1\times 10^{34}J$
0%
$2\times 10^{-50}J$

Explanation

The energy of photon is

$E=h\nu=\dfrac{hc}{\lambda}=\dfrac{(6.6\times 10^{-34})(3\times 10^8)}{2\times 10^{-9}} =9.9\times 10^{-17} J\sim 10^{-16} J$

Thus, the option C will correct.

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 5 - 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 5 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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