MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 13

The wave nature of particles was studied using diffraction of particle beams by crystal lattices. The wavelength of the waves associated with fast moving particles was found to be in agreement with the de Broglie relation.

For a particle of mass

$m$ moving with kinetic energy

$E$ , the de Broglie wavelength is

0%
$\displaystyle \dfrac{h}{2mE}$
0%
$\displaystyle h\sqrt{2mE}$
0%
$\displaystyle \dfrac{h}{\sqrt{2mE}}$
0%
$\displaystyle h\sqrt{\dfrac{2}{mE}}$

Explanation

The energy of photon is

$E = h\nu$
According to Einstein's energy mass relation

$E = mc^2$
From above equations

$h\nu = mc^2$

$\dfrac{hc}{\lambda} = pc$ ..................(since,

$p = mc$ )

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

This is the De-Broglie's equation for wavelength for the photon.
For the particle having mass m and velocity v, De-Broglie's equation for wavelength becomes

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

The energy of the particle is

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

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

$p = \sqrt {2mE}$
using this value in (1), we get

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

So, the answer is an option (C).

A photon of wavelength

$0.1\overset {o}{A}$ is emitted by a helium atom as a consequence of the emission of photon. The K.E. gained by helium atom is :

0%
0.05eV
0%
1.05 eV
0%
2.05 eV
0%
3.05 eV

Explanation

Given :

$\lambda = 0.1$

$A^o$

Momentum gained by atom

$P = \dfrac{h}{\lambda} = \dfrac{6.626\times 10^{-34}}{0.1 \times 10^{-10}} = 6.626\times 10^{-23}$

$kg.m/s$

Kinetic energy gained

$K.E = \dfrac{P^2}{2m} = \dfrac{(6.626\times 10^{-23})^2}{2(4\times 1.667\times 10^{-27})} =3.29\times 10^{-19}$ J

$\implies$

$K.E = \dfrac{3.29\times 10^{-19}}{1.6\times 10^{-19}} = 2.05$ eV

De Broglie wavelength of neutrons in thermal equilibrium is (given

$m_n=1.6\times 10^{-27} kg)$

0%
$30.8/\sqrt T\,\overset {o}{A}$
0%
$3.08/\sqrt T\,\overset {o}{A}$
0%
$0.308/\sqrt T\,\overset {o}{A}$
0%
$0.0308/\sqrt T\,\overset {o}{A}$

Explanation

The de Broglie wavelength,

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

$h=$ Planck's constant,

$m=$ mass of the particle ,

$k=$ Boltzmann's constant,

$T=$ temperature.

So,

$\lambda=\dfrac{6.62\times 10^{-34}}{\sqrt{2\times (1.6\times 10^{-27})\times (1.38\times 10^{-23})T}}=31.5\times 10^{-10} m/\sqrt T \sim 30.8/\sqrt T A^o$

A particle of mass

$M$ at rest decays into two particles of masses

$m _ { 1 } \text { and } m _ { 2 }$ having non zero velocities. The ratio of the de-Broglie wavelengths of the particles

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

0%
$m _ { 1 } / m _ { 2 }$
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$m _ { 2 } / m _ { 1 }$
0%
1.0
0%
$\sqrt { m _ { 2 } } / \sqrt { m _ { 1 } }$

Which of the following graphs correctly represents the variation of particle momentum with associated de Broglie wavelength?

Explanation

From de Broglie hypothesis, the momentum of a particle ,

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

$h=$ Plank's constant and

$\lambda=$ wavelength.

So the momentum versus wavelength graph is like

$y=1/x$ graph, which is represented in graph D.

$\alpha-particle$ and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The

$\alpha-particle$ and the proton move such that radius of curvature of their paths is same. Find the ratio of their de Broglie wavelengths :

0%
$2:3$
0%
$3:4$
0%
$5:7$
0%
$1:2$

Explanation

Magnetic force,

$\vec{F_B}=q(\vec{v}\times \vec B)=qvB\sin\theta$

As magnetic field is perpendicular to velocity so

$\theta=90^o$ so

$F_B=qvB$

or,

$\dfrac{mv^2}{r}=qvB$ or

$mv=qBr$

So, momentum

$p=mv=qBr$

de Broglie wavelength for proton ,

$\lambda_p=\dfrac{h}{p_{p}}=\dfrac{h}{q_pBr_p}$ and

de Broglie wavelength for alpha particle ,

$\lambda_{\alpha}=\dfrac{h}{p_{\alpha}}=\dfrac{h}{q_{\alpha}Br_{\alpha}}$

Given,

$\dfrac{r_{\alpha}}{r_p}=1$ and

$q_{\alpha}=2q_p$

Thus,

$\dfrac{\lambda_{\alpha}}{\lambda_p}=\dfrac{r_pq_p}{r_{\alpha}q_{\alpha}}=1/2$

A modern 200 W sodium street lamp emits yellow light of wavelength

$0.6\mu m$ . Assuming it to be 25% efficient in 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

Energy converted into light per second

$E = 0.25\times 200 = 50\ J$ per second

Given :

$\lambda = 0.6\mu m$

Energy of one photon

$E_p = \dfrac{hc}{\lambda} = \dfrac{6.63\times 10^{-34}\times 3\times 10^8}{0.6\times 10^{-6}} = 33.15\times 10^{-20} \ J$

Number of photons

$N = \dfrac{E}{E_p} = \dfrac{50}{33.15\times 10^{-20}} = 1.5\times 10^{20}$ photons per second

The human eye can barely detect a yellow light

$(\lambda =6000\overset {o}{A})$ that delivers

$1.7\times 10^{-18}$ W to the retina. The number of photons per second falling on the eye is nearest to

0%
$5\times 10^9$
0%
5000
0%
50
0%
5

Explanation

Energy of the yellow light is

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

Here the total energy per sec is given

$1.7\times 10^{-18} W$

Thus, number of photons per sec

$=\dfrac{1.7\times 10^{-18}}{3.31\times 10^{-19}}\sim 5$

A homogeneous ball

$(mass=m)$ of ideal black material at rest is illuminated with a radiation having a set of photons

$(wavelength=\lambda)$ , each with the same momentum and the same energy. The rate at which photons fall on the ball is n. The linear acceleration of the ball is:

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$m\lambda / nh$
0%
$nh/m\lambda$
0%
$nh/(2\pi)(m\lambda)$
0%
$2pm\lambda / nh$

Explanation

According to de-Broglie relation,

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

p = linear momentum = $m \times velcoity$

for 1 photon,

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

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

Since n= no. of photons fall on ball pers sec. and

Linear acceleration of the ball = Rate of change of velocity

So, $v = n \times \dfrac{h}{m\lambda}$

The eye can detect

$5\times 40^4$ photons

$(m^2s)^{-1}$ of green light

$(\lambda=5000\overset {o}{A})$ , while ear can detect

$10^{-13}W m^{-2}$ . As a power detector, which is more sensitive and by what factor?

0%
Eye is more sensitive and by a factor of 5.00
0%
Ear is more sensitive by a factor of 5.00
0%
Both are equally sensitive
0%
Eye is more sensitive by a factor of $10^{-1}$

Explanation

According to the given information the eye can detect the intensity

$I=nhv$

$=n\cfrac { hc }{ \lambda }$

$=\cfrac { 5\times { 10 }^{ 4 }\times 6.63\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 5\times { 10 }^{ -7 } }$

$\approx 0.2\times { 10 }^{ -13 }{ W }/{ ㎡ }$

$\therefore$ Eye can detect

$5$ times smaller intensities than that of ear. So, Eye

$5$ times more sensitive than ear.

An electron of mass

$m_e$ and a proton of mass

$m_p$ are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an electron to that associated with proton is

0%
1
0%
$m_p /m_e$
0%
$m_e/m_p$
0%
$\sqrt {m_p/m_e}$

Explanation

The relation between de Broglie wavelength

$(\lambda)$ and potential difference

$(V)$ is given by,

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

$V$ is constant so

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

Thus,

$\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}$

A sensor is exposed for time t to a lamp of power P placed at a distance l. The sensor has an opening that is 4d in diameter. Assuming all energy of the lamp is given off as light, the number of photons entering the sensor if the wavelength of light is

$\lambda$ is

0%
$N=P\lambda d^2t/hcl^2$
0%
$N=4P\lambda d^2t/hcl^2$
0%
$N=P\lambda d^2t/4hcl^2$
0%
$N=P\lambda d^2t/16hcl^2$

Explanation

Power is given by:

$P=\dfrac{E}{t}=\dfrac{nhc}{\lambda t}$

$n=\dfrac{P\lambda t}{hc}$

where

$n$ is the total number of photons emitted by the source.
The number of photons entering the sensor will be in proportion to the area of the sensor relative to the bigger sphere of radius

$l$ .
Thus, the number of photons entering the sensor is given by

$N=\dfrac{P\lambda t}{hc}\times \dfrac{\pi{(2d)}^{2}}{4\pi{l}^{2}}$

Thus,

$N=\dfrac{P\lambda {d}^{2}t}{hc{l}^{2}}$

The kinetic energy of a particle is equal to the energy of a photon. The particle moves at 5% of the speed of light. The ratio of the photon wavelength to the de Broglie wavelength of the particle is
[No need to used relativistic formula for the particle.]

0%
$40$
0%
$4$
0%
$2$
0%
$80$

A laser used to weld detached retinas emits light with a wavelength of

$652nm$ in pulses that are

$20.0ms$ in duration. The average power during each pulse is

$0.6 W$ . Then

0%
The energy of each photon is $3.048\times 10^{19} J$
0%
The energy content in each pulse is $12 mJ$
0%
The number of photons in each pulse is nearly $4\times 10^{15}$
0%
The energy of each photon is nearly $1.9 eV$

Explanation

Energy of a photon is given by:

$E=\dfrac{hc}{\lambda}=\dfrac{6.63\times{10}^{-34}\times3\times{10}^{8}}{652\times{10}^{-9}}$

$E=3.048\times{10}^{-19}J$ Thus, option A is incorrect.
Energy content in each pulse is given by

${E}_{p}=Power\times20\times{10}^{-3}=12mJ$ Thus, option B is correct.
Number of photons can be obtained using

$n=\dfrac{12\times{10}^{-3}}{3.048\times{10}^{-19}}=3.94\times{10}^{16}$ Thus, option C is also incorrect.

Energy of each photon in eV is given as

$E=3.048\times{10}^{-19}\times6.24\times{10}^{18} eV=1.9eV$ Thus, option D is correct.

A 100 W point source emits monochromatic light of wavelength

$6000\overset {o}{A}$ . Calculate the photon flux (in SI unit) at a distance of 5 m from the source. Given

$h=6.6\times 10^{34}$ J s and

$C=3\times 10^8 ms^{-1}$

0%
$10^{15}$
0%
$10^{18}$
0%
$10^{20}$
0%
$10^{22}$

Explanation

The photon flux is defined as the number of photons per second per unit area.

Thus, photon flux

$\phi=$ number of photons per second / area

$=\dfrac{n}{A}$

Energy of photon

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

Given, the energy per second is

$100 J$

Thus number of photons per sec,

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

Area

$A=4\pi\times 5^2=314 m^2$

Thus,

$\phi=\dfrac{3.02\times 10^{20}}{314}=9.6\times 10^{17} \sim 10^{18}$

The light sensitive compound on most photographic films is silver bromide AgBr. A film is exposed when the light energy absorbed dissociates this molecule into its atoms. The energy of dissociation of AgBr is

$10^5 J mol^{-1}$ . For a photon that is just able to dissociate a molecule of AgBr, the photon energy is

0%
1.04 eV
0%
2.08 eV
0%
3.12 eV
0%
4.16 eV

Explanation

1 mole of AgBr contains

$6.022\times{10}^{23}$ molecules and

$1J$ of energy is equivalent to

$6.24\times {10}^{18}eV$ .
Thus, energy needed to dissociate 1 molecule of AgBr is given by:

$E=\dfrac{ {10}^{5} \times {6.24\times {10}^{18} }}{6.022\times {10}^{23}} =1.04eV$

Determine the number of photons emitted by the laser each second.

0%
$3.18\times 10^{15}$
0%
$4.5\times 10^{16}$
0%
$1.2\times 10^{15}$
0%
$2.9\times 10^{17}$

Explanation

The energy of a photon is

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

$h=$ Planck's constant,

$c=$ velocity of light.

So,

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

Total power 1 mW means the energy per sec is given

$1 m J=10^{-3} J$

Thus, number of photons emitted per sec

$=\dfrac{10^{-3}}{3.14\times 10^{-19}}=3.18\times 10^{15}$

A photon has same wavelength as the de Broglie wavelength of electrons. Given

$C=$ speed of light,

$v=$ speed of electron. Which of the following relation is correct? [Here

$E_e=$ kinetic energy of electron,

$E_{ph}=$ energy of photon,

$P_e=$ momentum of electron and

$P_{ph}=$ momentum of photon]

0%
$E_e/E_{ph}=2C/v$
0%
$E_e/E_{ph}=v/2C$
0%
$P_e/P_{ph}=2C/v$
0%
$P_e/P_{ph}=C/v$

Explanation

Given,

$\lambda_{ph}=\lambda_e ....(1)$

Now,

$\lambda_e=\dfrac{h}{P_e}=\dfrac{h}{mv} ....(2)$

Energy of photon,

$E_{ph}=\dfrac{hC}{\lambda_{ph}}=\dfrac{hC}{\lambda_e}$ (using (1))

Energy of electron,

$E_e=\dfrac{P_e^2}{2m}=\dfrac{h^2}{\lambda_e^2(2m)}$ using (2)

Now,

$\dfrac{E_e}{E_{ph}}=\dfrac{h^2}{\lambda_e^2(2m)}\times \dfrac{\lambda_e}{hC}=(1/2mC)(h/\lambda_e)=(1/2mC)(mv)=v/2C$ using (2)

Now,

$\dfrac{P_e}{P_{ph}}=\dfrac{h/\lambda_e}{h/\lambda_{ph}}=1$ using (1)

Two identical non-relativistic particles A and B move at right angles to each other, processing de Broglie wavelengths

$\lambda_1$ and

$\lambda_2$ respectively. The de Broglie wavelength of each particle in their centre of mass frame of reference is :

0%
$\lambda_1+\lambda_2$
0%
$2\lambda_1\lambda_2/(\sqrt {\lambda_1^2+\lambda_2^2})$
0%
$\lambda_1\lambda_2/(\sqrt {|\lambda_1^2+\lambda_2^2|})$
0%
$(\lambda_1+\lambda_2)/2$

Explanation

Two identical non-relativistic particles A and B move at right angles to each other with say velocity

$v_1, v_2$ . along X and Y directions respectively (say).Therefore Velocity of center of mass (COM) ;

$v_c= v_1\hat i+v_2\hat j$ Velocity of 1st particle with respect to COM;

$v_{1c}=v_1-v_c=\dfrac{v_1\hat i-v_2 \hat j}{2}$
Velocity of 2nd particle with respect to COM;

$v_{2c}=v_2-v_c=\dfrac{v_2\hat j-v_1 \hat i}{2}$ De Broglie wavelength of 1st particle w.r.t COM

$\lambda_{1c}=\dfrac{h}{mv_{1c}}=\dfrac{2h}{m\sqrt{v_1^2+v_2^2}}$

$\lambda_{1c}=\dfrac{2h}{m\sqrt{(\dfrac{h}{m\lambda_1})^2+(\dfrac{h}{m\lambda_2})^2}}$

$\lambda_{1c}=\dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$ Similarly,De Broglie wavelength of 2nd particle w.r.t COM

$\lambda_{2c}=\dfrac{h}{mv_{2c}}=\dfrac{2h}{m\sqrt{v_1^2+v_2^2}}$

$\lambda_{2c}=\dfrac{2h}{m\sqrt{(\dfrac{h}{m\lambda_1})^2+(\dfrac{h}{m\lambda_2})^2}}$

$\lambda_{2c}=\dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$

The human eye is most sensitive to green light of wavelength 505 nm. Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina. The velocity of typical bacterium of mass

$9.5\times 10^{-12}$ g, if it had absorbed all energy of photon, is nearly:

0%
$9\times10^{-3} ms^{-1}$
0%
$10^{8} ms^{-1}$
0%
$10^{-10} ms^{-1}$
0%
$10^{-13} ms^{-1}$

Explanation

If the bacterium absorbs photon

$\dfrac { 1 }{ 2 } m{ v }^{ 2 }=\dfrac { hc }{ \lambda } \\ \lambda =505\times { 10 }^{ -9 }m\\ m=9.5\times { 10 }^{ -3 }\times { 10 }^{ -12 }kg\\ v=\sqrt { \frac { hc }{ m\lambda } } =\sqrt { \dfrac { 2\times 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 9.5\times { 10 }^{ -15 }\times 505\times { 10 }^{ -9 } } } \\ =9\times { 10 }^{ -3 }m/sec\\$

Two electrons are moving with same speed

$v$ . One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, when after some time de Broglie wavelengths of two are

$\lambda_1$ and

$\lambda_2$ , respectively. Now :

0%
$\lambda_1 < \lambda_2$
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$\lambda_1=\lambda_2$
0%
$\lambda_1 > \lambda_2$
0%
$\lambda_1$ can be greater than or less than $\lambda_2$

Explanation

In a uniform electric field, an electron has a motion with uniform acceleration in the direction of the field with a magnitude which equals

$\dfrac{eE}{m_e}$ whereas in a uniform magnetic field, an electron undergoes a circular motion with the centripetal force of magnitude

$evB$ . Since the motion is circular in the magnetic field the magnitude of the velocity of the electron will be equal to the initial velocity of

$v$ .

$\therefore$

$\lambda_2 = \dfrac{h}{mv}$

But in the case of the uniform electric field, the final velocity depends on the direction of the electric field and the time period for which it is applied. If the applied field is opposite to the initial direction then the final velocity could be zero which means

$\lambda_1 > \lambda_2$ whereas if the applied field is in the direction of initial motion then

$\lambda_1 < \lambda_2$ after some time.

The magnitude of the de-Broglie wavelength

$(\lambda)$ of electron (e), proton (p), neutron (n) and

$\alpha$ -particle

$(\alpha)$ all having the same energy of 1MeV, in the increasing order will follow the sequence

0%
$\lambda _e, \lambda _p, \lambda _n, \lambda _\alpha$
0%
$\lambda _e, \lambda _n, \lambda _p, \lambda _\alpha$
0%
$\lambda _\alpha , \lambda _n, \lambda _p, \lambda _e$
0%
$\lambda _p , \lambda _e, \lambda _\alpha , \lambda _n$

Explanation

$E = \cfrac{1}{2}mv^2$

$p=mv$

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

$p = \sqrt{2mE}$
De-Broglie wavelength

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

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

Given Energy is same for all

$\lambda \sqrt{m} = const.$

$\lambda \propto \cfrac{1}{\sqrt{m}}$

Hence, lower the mass, higher the wavelength

$m_{\alpha} > m_{n}> m_p > m_e$

$\lambda_{\alpha} < \lambda_{n}< \lambda_p <\lambda_e$

$10,000V$ are applied across an X-ray tube, find the ratio of wavelength of the incident electrons and the shortest wavelength of X-ray coming out of the X-ray tube, given

$e/m$ of electron

$=1.8\times10^{11}\space C\space kg^{-1}$ .

0%
$1:10$
0%
$10:1$
0%
$5:1$
0%
$1:5$

Explanation

$V=10kV=10^4\ volts$

The shortest wavelength is given as,

$\lambda_1=hc/eV_o$

The de-broglie wavelength is given as

$\lambda_2=h/p=h/\sqrt{2\times eV_o\times m_e}$ wher

$m_e$ is the mass of the electron,

$p$ is the momentum.

We can further write de-broglie wavelength as

$\lambda_2=hc/(eV_o\sqrt{2V_oc^2/\dfrac{e}{m_e}})=hc/10eV_o$

$\Rightarrow \lambda_2=\lambda_1/10$

So ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced is

$\lambda_2:\lambda_1=1:10$

The electron cannot exist inside the nucleus because

0%
de-Broglie wavelength associated with electron in $\beta$ -decay is much less than the size of nucleus
0%
de-Broglie wavelength associated with electron in $\beta$ -decay is much greater than the size of nucleus
0%
de-Broglie wavelength associated with electron in $\beta$ -decay is equal to the size of nucleus
0%
de-Broglie negative charge cannot exist in the nucleus

Explanation

${ \beta }^{ - }\quad decay$ :

${ \beta }^{ - }$ radioactivity occurs when a nucleus emits a negative electron from an unstable radioactive nucleus whose deBroglie wavelength of order

$MeV$ is much larger than nuclear dimensions.

The number of photons that each pulse delivers to the blemish is

0%
$1.5\times 10^{16}$
0%
$1.5\times 10^{8}$
0%
$3\times 10^{16}$
0%
$3\times 10^{8}$

Explanation

Number of photons

$=\upsilon =\dfrac { C }{ \lambda }$

$=\dfrac { 3\times { 10 }^{ 8 }\times { 10 }^{ 9 } }{ 585 }$

$=5.12\times { 10 }^{ 14 }$

Number of photons

$=0.05\times { 10 }^{ 16 }\simeq$ Closer to

$1.5\times { 10 }^{ 16 }$

The circumference of the second Bohr orbit of electron in hydrogen atom is

$600nm$ . The potential difference that must be applied between the plates so that the electrons have the de Broglie wavelength corresponding in this circumference is

0%
$10^{-5}\space V$
0%
$\displaystyle\frac{5}{3}\times10^{-5}\space V$
0%
$5\times10^{-5}\space V$
0%
$3\times10^{-5}\space V$

Explanation

For second Bohr orbit,

$600nm=n\lambda =2\lambda \Rightarrow \lambda =3000{ A }^{ o }$

For electron

$\lambda =\dfrac { 12.27 }{ \sqrt { V } } { A }^{ o }$

Voltage

$V={ \left( \dfrac { 12.27 }{ 3000 } \right) }^{ 2 }=\dfrac { 150.553 }{ 9 } \times { 10 }^{ -6 }=1.67\times { 10 }^{ -5 }=\dfrac { 5 }{ 3 } \times { 10 }^{ -5 }V$

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 :

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

Explanation

By conservation of momentum,

$Mv=m_1v_1+m_2v_2$

As mass M is at rest so

$v=0$ . Thus,

$m_1v_1=-m_2v_2$ , it indicates the magnitude of momentum of masses is same i.e

$p_1=p_2$

So, de Broglie wavelengths for masses

$m_1$ and

$m_2$ are

$\lambda_1=h/p_1$ and

$\lambda_2=h/p_2$

Thus,

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

$(p_1=p_2)$

The energy of the photons causing the photoelectric emission is

0%
$2.55\space eV$
0%
$0.73\space eV$
0%
$1.82\space eV$
0%
Information insufficient

Explanation

Given, work function for Sodium is

$1.82eV$ and

$K.E$ of fastest electron emitted is

$0.73eV$ .
Total energy incident on Sodium is

$2.55eV.$

The de-Broglie wavelength of a neutron at

$972^o C$ is

$\lambda$ .
What will be its wavelength at

$27^o C$ ?

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

Explanation

$E = \cfrac{1}{2}mv^2 = \cfrac{p^2}{2m}$

De - Broglie wavelength

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

Now for a molecule

$E = KT$ (Boltzmann Law)

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

$\lambda = \cfrac{C}{\sqrt{T}}$

$\cfrac{\lambda_1}{ \lambda_2} = \sqrt{\cfrac{T_2}{T_1}}$

Here T is in K

$\cfrac{\lambda_1}{ \lambda_2} = \sqrt{\cfrac{300}{1245}}$

$\lambda _2 = 2 \lambda_1$

A steel ball of mass m is moving with a kinetic energy 'K'. The de-Broglie wavelength associated with the ball is

0%
$\dfrac{h}{2mK}$
0%
$\sqrt{\dfrac{h}{2 mk}}$
0%
$\dfrac{h}{\sqrt{2mK}}$
0%
Meaningless

Explanation

$E = \cfrac{1}{2}mv^2$

$p=mv$

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

$p = \sqrt{2mE}$
De-Broglie wavelength

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

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

Two insulating plates are both uniformly charged in such a way that the potential difference between them is

$V_2-V_1=20 V$ . (i.e., plate 2 is at a higher potential). The plates are separated by

$d=0.1$ m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate

$1$ . What is its speed when it hits plate

$2$ ?

$(e=1.6 \times 10^{-19} C, m_e=9.11 \times 10^{-31} kg)$

0%
$2.65 \times 10^6 m/s$
0%
$7.02 \times 10^{12} m/s$
0%
$1.87 \times 10^6 m/s$
0%
$32 \times 10^{-19} m/s$

Explanation

The energy by accelerating the electron is in the form of kinetic energy.

So,

$Ve = \cfrac{1}{2}mv^2$

$v = \sqrt{\cfrac{2Ve}{m}}$

$v =\sqrt{\cfrac{2 \times 20 \times (1.6 \times 10^{-19})}{9.1 \times 10^{-31}}}$

$v =2.65 \times 10^6 m/s$

In case of electrons and photons having the same wavelength. What is same for them ?

0%
Energy
0%
Velocity
0%
Momentum
0%
Angular momentum

Explanation

For electron, Debroglie wavelength

$\lambda _e = \cfrac{h}{p}$

For Photon,

$\lambda _p = \cfrac{h}{p}$
Hence equating, momentum has to be same for both of them to have same wavelength

If 5% of the energy supplied to a bulb is radiated as visible light, the number of visible quanta emitted per second by a 100W bullb, assuming the wavelength of visible light to be

$5.6 \times 10^{-5}cm$ , is

0%
$1.4 \times 10^{19}$
0%
$1.4 \times 10^{20}$
0%
$2 \times 10^{19}$
0%
$2 \times 10^{20}$

Explanation

Energy of single photon =

$\cfrac{hc}{\lambda}$

If there are

$n$ photons, total energy =

$\cfrac{nhc}{\lambda}$

Total energy emitted =

$0.05 \times 100 W = 5 W/s$
Equating

$\cfrac{nhc}{\lambda} = 5$

$n = \cfrac{5 \lambda}{hc}$

Substitute values

$n = \cfrac{5 \times 5.6 \times 10^{-7}}{6.26 \times 10^{-34} \times (3 \times 10^8)}$

$n= 1.4 \times 10^{19}$

For a photoelectric cell, the graph showing the variation of cut off voltage

$(V_0)$ with frequency

$(v)$ of incident light is:-

Explanation

[C] is the correct option.

According to Einstein's equation

$hv={ W }_{ 0 }+{ K }_{ max }\Rightarrow { V }_{ 0 }=\left( \dfrac { h }{ c } \right) v-\dfrac { { W }_{ 0 } }{ e }$

This is the equation of straight line having positive slope

$(h/e)$ and intercept on

$-{ V }_{ 0 }\dfrac { { W }_{ 0 } }{ c }$ .

The de-Broglie wavelength of a proton

$(mass=1.6 \times 10 ^{-27} kg)$ accelerated through a potential difference of 1 kV is :

0%
$600 \mathring {A}$
0%
$0.9 \times 10^{-12}m$
0%
$7 \mathring { A }$
0%
$0.9 \times 10^{-19}nm$

Explanation

$eV = KE = \cfrac{1}{2}mv^2$

$KE = \cfrac{p^2}{2m}$

De-broglie wavelength,

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

$\lambda = \cfrac{h}{\sqrt{2mE}} = \cfrac{h}{\sqrt{2mVe}}$

Putting the values of constants, we get:

$\lambda = \dfrac {6.6 \times 10^{-34}}{\sqrt {2 \times 1.6 \times 10^{-27} \times 1.6 \times 10^{-19} \times 1000 } }= 0.9 \times 10^{-12} m$

A material particle with a rest mass

$'m_0\ '$ is moving with speed of light 'c'. The de-Broglie wavelength associated is given by

0%
$\dfrac{h}{m_0c}$
0%
$\dfrac{m_0c}{h}$
0%
Zero
0%
$\infty$

Explanation

Theory of special relativity

$m' = \gamma m = \cfrac{m}{\sqrt{1 -\cfrac{V^2}{c^2}}}$

Debroglie wavelength

$\lambda = \cfrac{h}{m'v} = \cfrac{h \sqrt{1 -\cfrac{V^2}{c^2}}}{mv}$

$v=c$
then

$\lambda =0$

A monochromatic source of light operating at 200 W emits

$4 \times 10^{20}$ photons per second. Find the wavelength of light.

0%
$400 nm$
0%
$200 nm$
0%
$4 \times 10 ^{-10} \mathring { A }$
0%
None of these

Explanation

Energy of single photon =

$\cfrac{hc}{\lambda}$
If there are

$n$ photons, total energy =

$\cfrac{nhc}{\lambda}$
Total energy emitted =

$200 W$
Equating

$\cfrac{nhc}{\lambda} = 200$

$n = \cfrac{200 \lambda}{hc}$

Substitute values:

$4 \times 10^{20} = \cfrac{200 \times \lambda}{6.26 \times 10^{-34} \times (3 \times 10^8)}$

$\lambda = 400 nm$

A particle with rest mass

$'m_0\ '$ is moving with speed of light 'c'. The de-Broglie wavelength associated with it will be

0%
$\infty$
0%
Zero
0%
$m_0 \, c/h$
0%
$hv/m_0c$

Explanation

Theory of special relativity,

$m' = \gamma m = \cfrac{m}{\sqrt{1 -\cfrac{V^2}{c^2}}}$

Debroglie wavelength,

$\lambda = \cfrac{h}{m'v} = \cfrac{h \sqrt{1 -\cfrac{V^2}{c^2}}}{mv}$
If

$v=c$ , then

$\lambda =0$

$E_1, E_2, E_3$ are the respective kinetic energies of an electron, an

$\alpha$ -particle and a proton, each having the same de-Broglie wavelength, then

0%
$E_1 > E_3 > E_2$
0%
$E_2 > E_3 > E_1$
0%
$E_1 > E_2 > E_3$
0%
$E_1 = E_2 = E_3$

Explanation

$KE = \cfrac{1}{2}mv^2$

$KE = \cfrac{p^2}{2m}$
Debroglie wavelength

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

$\lambda = \cfrac{h}{\sqrt{2mE}}$
Given

$\lambda_1 = \lambda_2 = \lambda_3$

$m_1E_1 = m_2E_2 = m_3E_3$
Since

$m_1<m_3<m_2$

$E_1>E_3>E_2$

The de - Broglie wavelength associated with the electron in the

$n=4$ level is :

0%
half of the de-Broglie wavelength of the electron in the ground state
0%
four times the de-Broglie wavelength of the electron in the ground state
0%
$1/4^{th}$ of the de-Broglie wavelength of the electron in the ground state
0%
two times the de-Broglie wavelength of the electron in the ground state

Explanation

De-Broglie wavelength is given as

$\lambda =\dfrac{h}{mv}$ ...(i) where h is planck constant

The velocity of electron in Bohr model is given as

$v=\dfrac{h}{2\pi m_e a_\circ{} n}$ ...(i)

in the ground state n=1

$v_{n=1}=\dfrac{h}{2\pi m_e a_\circ{} }$

$v_{n=4}=\dfrac{h}{2\pi m_e a_\circ{} 4}$

$v_{n=4}=\dfrac{v_{n=1}}{4}$ ...(iii)

from equation (i)

$\lambda_{n=4}=\dfrac{h}{m_e v_{n=4} }$

substituting value of

$v_{n=4}$ from equation (iii)

$\lambda_{n=4}=\dfrac{h}{m_e \dfrac{ v_{n=1}}{4} }$

$\lambda_{n=4}=4 \lambda_{n=1}$

Hence correct answer will be option B.

Saturation current in second case is

0%
$\displaystyle 9.2\mu A$
0%
$\displaystyle 13.3\mu A$
0%
$\displaystyle 15.1\mu A$
0%
$\displaystyle 19.4\mu A$

Explanation

Given:

$\lambda_2 = 1650 A^o$ Power

$P_2 = 5 mW$

Rate of incident photons

$R_2 = \dfrac{Power}{Energy} = \dfrac{P_2}{E_2} = \dfrac{P_2 \lambda_2}{hc}$

$\therefore R'_p = \dfrac{5 \times 10^{-3} \times 1650 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 4.17 \times 10^{15}$ photons per sec

As the photoelectron generation per incident photon

$\eta = 2$ % (solved earlier)

$\therefore$ Rate of electron emitted

$R'_e = \eta \times R'_p = \dfrac{2}{100} \times 4.17 \times 10^{15} = 8.34 \times 10^{13}$ electrons per sec

$\therefore$ Saturation current

$I'_s = 8.34 \times 10^{13} \times 1.6 \times 10^{-19} = 13.34$

$\mu A$

Each question contains statement 1 and statementChoose the correct answer (only one option is correct) from the following options.

Statement-1 : The de-Broglie wavelength of a molecule (in a sample of ideal gas) varies inversely as the square root of absolute temperature.
Statement-2 : The rms velocity of a molecule (in a sample of ideal gas) depends on temperature.

0%
Statement-1 is false, Statement -2 is true
0%
Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for statement-1
0%
Statement -1 is true, Statement-2 is true; Statement -2 is not correct explanation for statement-1
0%
Statement-1 is true, Statement -2 is false

Explanation

$v_{rms} = \sqrt{\dfrac{3RT}{M}}$

$v_{rms} \propto \sqrt{T}$

De broglie wavelength

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

$\lambda \propto \cfrac{1}{v_{rms}}$

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

If velocity of a particle A is 50% of velocity of particle B and mass of B is 25% of mass of A then de-Broglie wavelength of B if wavelength of A is

$2\mathring A$

0%
$4A^{\circ}$
0%
$20A^{\circ}$
0%
$6.0A^{\circ}$
0%
none

An LED (Light Emitting Diode) is constructed from a p-n junction based on a certain

$Ga-As-P$ semi-conducting material whose energy gap is

$1.9eV$ .What is the wavelength of the emitted light?

0%
$650nm$
0%
$65\mathring { A }$
0%
$800nm$
0%
$8000\mathring { A }$

Explanation

The wavelength of emitted light

$\lambda=\cfrac{hc}{{E}_{g}}$
Where

${E}_{g}=$ energy gap of semiconductor

$=1.9eV$

$=1.9\times 1.6\times {10}^{-19}V$

$\lambda=\cfrac{6.6\times {10}^{-34}\times 3\times {10}^{8}}{1.9\times 1.6\times {10}^{-19}}m$

$6.5\times {10}^{-7}m$

$=650\times {10}^{-9}m$

$=650nm$

The de-Broglie wavelength of an electron (mass

$1\times { 10 }^{ -30 }kg$ , charge

$=1.6\times { 10 }^{ -19 }\ C$ ) with kinetic energy of

$200eV$ is: (Planck's constant

$6.6\times { 10 }^{ -34 }J$ ):

0%
$9.60\times { 10 }^{ -11 }m$
0%
$8.25\times { 10 }^{ -11 }m$
0%
$6.25\times { 10 }^{ -11 }m$
0%
$5.00\times { 10 }^{ -11 }m$

Explanation

The de-Broglie wavelength

$\lambda =\cfrac { h }{ \sqrt { 2mk } }$
Given

$h=6.6\times { 10 }^{ -34 }Js$

$m=1\times { 10 }^{ -30 }kg$

$K=200eV=200\times 1.6\times { 10 }^{ -19 }\quad J$
Substituting all these values

$\lambda =\cfrac { 6.6\times { 10 }^{ -34 } }{ \sqrt { 2\times 1\times { 10 }^{ -30 }\times 200\times 1.6\times { 10 }^{ -19 } } }$

$=0.825\times { 10 }^{ -10 }=8.25\times { 10 }^{ -11 }m$

A hydrogen atom makes a transition from

$n=3$ to

$(n=1)$ .The momentum of the emitted photon is $${h=Planck' constant,R=Rydberg constant)

0%
$\frac { 8hR }{ 9 }$
0%
$\frac { 9hR }{ 9 }$
0%
$\frac { 8h }{ 9 }$
0%
$\frac { 8R }{ 9 }$

When a microgram of matter is converted to energy, the amount of energy released will be?

0%
$9\times 10^{14}$ J
0%
$9\times 10^{10}$ J
0%
$9\times 10^7$ J
0%
$9\times 10^4$ J

The de-Broglie wavelength of an electron is the same as that of a

$50 keV$ X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is

$0.5 MeV$ ):

0%
$1 : 50$
0%
$1 : 20$
0%
$20 : 1$
0%
$50 : 1$

Explanation

Energy of X-ray photon

$E_{photon} = 50 keV = 5\times 10^4 eV$

Energy equivalent of electron mass is

$0.5MeV$ i.e

$mc^2 = 0.5 MeV =5 \times 10^5 eV$ ......(1)

Wavelength of the photon

$\lambda_{photon} = \dfrac{hc}{E} = \dfrac{1242}{5 \times 10^4}$

$nm = 0.0248$

$nm$

$\therefore$ de-Broglie wavelength of the electron

$\lambda = \lambda_{photon} = 0.0248$

$nm$

Kinetic energy of electron

$E_{electron} = \dfrac{h^2}{2m \times \lambda^2} = \dfrac{h^2c^2}{2 (mc^2) \lambda^2}$

$(\because E = \dfrac{p^2}{2m})$

Kinetic energy of electron

$E_{electron} = \dfrac{(1242)^2}{2 (5\times 10^5) \times (0.0248)^2}$

$eV = 2.51 \times 10^3 eV$

$\therefore$

$\dfrac{E_{photon}}{E_{electron}} = \dfrac{5 \times 10^4}{2.51 \times 10^3} =20$

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

$1.5\times 10^8\ m/s$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon

$(c=3\times 10^8\ m/s)$ :

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

Explanation

Since the de-Broglie wavelength depends only on the momentum of particle, both electron and photon must have same momentum. Also, kinetic energy expression can be written as,

$K=\dfrac { 1 }{ 2 } m{ v }^{ 2 }=\dfrac { 1 }{ 2 } pv$

Hence,

$\dfrac { { E }_{ electron } }{ { E }_{ photon } } =\dfrac { { v }_{ electron } }{ { v }_{ photon } } =\dfrac { 1.5 \times { 10 }^{ 8 } }{ 3 \times { 10 }^{ 8 } } =\dfrac { 1 }{ 2 }$

A proton when accelerated through a potential difference of

$V$ , has a de Broglie wavelength

$\lambda$ associated with it. If an alpha particle is to have the same de Broglie wavelength

$\lambda$ , it must be accelerated through a potential difference of:

0%
$\displaystyle\frac{V}{8}$
0%
$\displaystyle\frac{V}{4}$
0%
$4V$
0%
$8V$

Explanation

Since the kinetic energy will be equal to the work done on particle,

$K=qV$

Also,

$p=\sqrt { 2mK }$

Thus,

$\lambda =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mK } } =\dfrac { h }{ \sqrt { 2mqV } }$

For both to have same

$\lambda$ ,

${ m }_{ p }{ q }_{ p }{ V }={ m }_{ a }{ q }_{ a }{ V }_{ a }$

${ V }_{ a }=\dfrac { 1 }{ (4)(2) } V=\dfrac { V }{ 8 }$

Answer is option A.

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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