MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 10

The electron in the hydrogen atom jumps from the second orbit to the forth orbit after absorbing photon. In this process its:

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energy doubles
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angular momentum doubles
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velocity doubles
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linear momentum doubles

Photoelectrons are ejected from a metal when light of frequency v falls on it. Pick out the wrong statement from the following

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No electrons are emitted if v less than W/h , where W is the work function of the metal
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The ejected of the photoelectrons is instantaneous.
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The maximum energy of the photoelectrons is hv.
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The maximum energy of the photoelectrons is independent of the intensity of the light.

Explanation

The frequency of incident light is less than the threshold frequency. No photoelectrons will be emitted from the metal surface. The threshold frequency is $7.24 \times 1014 Hz,$ and no photoelectron will be ejected.

Which of the following statement is

$NOT$ correct for the saturation current in a photoelectric cell ?

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All the electrons emitted from the photosensitive plate reach collector.
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The potential difference between the emitter and collector should be numerically equal to the stopping potential.
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Collector is positive with respect to emitter.
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It is the maximum current that can be set up in the photoelectric cell.

For wave connected with proton, de-broglie wavelength change by 0.25% if its momentum change by

$P_0$ initial momentum=

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$100 P_0$
0%
$\dfrac {P_0}{400}$
0%
$401 P_0$
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$\dfrac {P_0}{100}$

Explanation

Using,

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

$\lambda =$ wavelength

$p=$ Momentum

While Reducing wavelength by

$0.25$ % then momentum reduces by

$1$ unit.

So,

$\left( 1-0.5\lambda \right) =\dfrac { h }{ p }$

Initial Momentum

$=$

$401{ p }_{ 0 }$

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to

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$n^2$
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n
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$n^{1/2}$
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$n^{1/4}$

The photoelectric effect can be understood on the basis of

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The principle of superposition
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The electromagnetic theory of light
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The special theory of relativity
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Line spectrum of the atom

Which of the following is an to a possible de-Broglie's wavelength of a particle, which moves inside a cubical box of side length

$L$ , without losing any energy (elasticity colliding with walls of cube)?

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0%
0%
0%
All of these

The circumstance of first of hydrogen atom is s.Then the Broglie wavelength of electron to that orbit is

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$\dfrac {S} {2}$
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$2S$
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$S$
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$3S$

Explanation

Broglie wavelength

$\lambda = n\left( {2\pi R} \right)$

$\lambda = S$

Hence,

option

$(c)$ is correct answer.

A proton and an

$\alpha$ particle are accelerated through the same potential difference V. The ratio of their de Broglie wavelengths is?

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$\sqrt{2}$
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$2\sqrt{2}$
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$\sqrt{3}$
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$2\sqrt{3}$

Explanation

$\lambda =\dfrac{h}{\sqrt{2mqV}}\Rightarrow \lambda \sqrt{mq}=com$

$\Rightarrow \dfrac{\lambda_p}{\lambda_{\alpha}}=\sqrt{\dfrac{2\times 4}{1\times 1}}=2\sqrt{2}$ .

A particle

$A$ of mass

$m$ and initial velocity

$v$ collides with a particle

$B$ of mass

$\dfrac {m}{2}$ which is at rest. The collision is held on, and elastic. The ratio of the de-Broglie wavelength

$\lambda_{A}$ to

$\lambda_{B}$ after the collision is

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$\dfrac {\lambda_{A}}{\lambda_{B}} = 2$
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$\dfrac {\lambda_{A}}{\lambda_{B}} = \dfrac {2}{3}$
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$\dfrac {\lambda_{A}}{\lambda_{B}} = \dfrac {1}{2}$
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$\dfrac {\lambda_{A}}{\lambda_{B}} = \dfrac {1}{3}$

Explanation

By conservation of linear momentum

$mV={ mV }_{ 1 }+\dfrac { m }{ 2 } { V }_{ 2 }$

$2V=2{ V }_{ 1 }+{ V }_{ 2 }\quad \longrightarrow \left( 1 \right)$

${ \lambda }_{ 1 }=\dfrac { h }{ { p }_{ 1 } }$ ;

${ \lambda }_{ 2 }=\dfrac { h }{ { O }_{ 2 } }$

By law of collision

$e=\dfrac { { V }_{ 2 }-{ V }_{ 1 } }{ { U }_{ 1 }-{ U }_{ 2 } }$

$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\dfrac { 2 }{ 1 }$

$u={ V }_{ 2 }-{ V }_{ 1 }\quad \longrightarrow \left( 2 \right)$

$\boxed { { \lambda }_{ 1 }:{ \lambda }_{ 2 }=2:1 }$

equ

$(1)$ and

$(2)$

${ V }_{ 1 }=\dfrac { V }{ 3 } \quad ;\quad { V }_{ 2 }=\dfrac { 4V }{ 3 }$

Light of wavelength

$4000\overset { 0 }{ A }$ is incident on a sodium surface for which the threshold wavelength of photoelectrons is

$5420\overset { 0 }{ A }$ . The work function of sodium is

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0.57 eV
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1.14 eV
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2.29 eV
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4.58 eV

Explanation

As we know,

$W=\dfrac {12375}{\lambda_0(A)}=\dfrac {12375}{5420}=2.28\ eV$

Figure represents the graph of photo-current

$I$ versus applied voltage

$(V)$ . The maximum energy of the emitted photoelectron is-

0%
$2eV$
0%
$4eV$
0%
$0eV$
0%
$3eV$

Explanation

We see that for voltage

$-4volt$ the current is

$zero$ so

$magnitude$ of the cutoff voltage will be

$V_0=4volt$

So maximum kinetic energy of emitted photoelectron will be

$eV_0=4eV$

Option B is correct.

Two large parallel plates are connected with the terminal of

$100V$ power supply. These plates have a fine hole at the centre. An electron having energy

$200eV$ is so directed that it passes through the holes. When it comes out it's de-Broglie wavelength is

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$1.22 A ^ { \circ }$
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$1.75 A ^ { \circ }$
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$2 A ^ { \circ }$
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none of these

Explanation

Energy of the electron, when it comes out from the second plate

$=200\ eV-100\ eV=100\ eV$

Hence acceleration potential difference

$=100\ V$

$\lambda_{Electron}=\dfrac{12.27}{\sqrt{V}}=\dfrac{12.27}{\sqrt{100}}=1.23\overset{o}{A}$

The de-Broglie wavelength of a proton accelerated by

$400\ V$ is

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$0.005\ \mathring {A}$
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$1.0528\ \mathring {A}$
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$0.0568\ \mathring {A}$
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$0.0143\ \mathring {A}$

Explanation

$0.043\dot{A}$

formulae,

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

$\Rightarrow v=\sqrt{\dfrac{2qV}{m}}$

$\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 400}{1.67\times 10^{-27}}}=2.77\times 10^{5}m/s$

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

$=\dfrac{6.63 \times 10^{-34}}{(1.67\times 10^{-27})(2.77\times 10^{5})}$

$\therefore \lambda =1.43\times 10^{-12}m$

$\Rightarrow \lambda =0.0143\dot{A}$

Two particles A and B have de-Broglie's wavelengths

$30 \ \mathring A$ and

$20 \ \mathring A$ , combined to form a particle C. Momentum is conserved in this process. The possible de-Broglie's wavelength of C is :

(the motion is one dimensional)

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$12 \ \mathring A$
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$20 \ \mathring A$
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$10 \ \mathring A$
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$22 \ \mathring A$

Explanation

Given,

$\lambda _A=30\times 10^{-10}m$

$\lambda _B=20\times 10^{-10}m$

Momentum is conserved, so by applying the conservation of momentum,

$P_C=P_A+P_B$

$\dfrac{h}{\lambda _C}=\dfrac{h}{\lambda _A}+\dfrac{h}{\Lambda _B}$

$\dfrac{1}{\lambda _C}=\dfrac{1}{30\times 10^{-10}}+\dfrac{1}{20\times 10^{-10}}$

$\lambda _C=12\times 10^{-10}m$

A proton and an electron are accelerated by same potential difference starting from rest have de- Brogile wavelength

$\lambda _ { p }$ and

$\lambda _ { e ^ { * } }$

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$\lambda _ { e } = \lambda _ { p }$
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$\lambda _ { 0 } < \lambda _ { p }$
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$\lambda _ { e } > \lambda _ { p }$
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none of these

Explanation

The wavelength of the photon is given as:

$\begin{array}{l} \lambda =\dfrac { h }{ P } \\ =\dfrac { h }{ { \sqrt { 2mk } } } \\ \lambda \propto \dfrac { 1 }{ { \sqrt { m } } } \\ \therefore { \lambda _{ p } }<{ \lambda _{ e } } \\ Hence, \ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

The number of photo electrons emitted for light of a frequency v (higher than the threshold frequency

$V_{0}$ ) is proportional to : -

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Frequency of light (v)
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$v - v_{o}$
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Threshold frequency $(v_{o})$
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Intensity of light

Electron microscope is based on the principle

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Photoelectric effect
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wave nature of electron
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Super conductivity
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Laws of electro magneitc induction

The energies of three coservatives energy levels

${ l }_{ 3 }$ ,

${ l }_{ 2 }$ and

${ l }_{ 1}$ of

${ E }_{ 0\quad }\dfrac { { 4E }_{ 0 } }{ 9 } and\quad \dfrac { { E }_{ 0 } }{ 4 }$ respectively. A photon of wavelength

$\lambda$

${ l }_{ 3 }\quad to\quad { l }_{}$ What will be the wavelength of emission for transition

${ l }_{ 3 }\quad to\quad { l }_{}$ ?

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$\dfrac { 16\lambda }{ 31 }$
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$\dfrac { 27\lambda }{ 7 }$
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$\dfrac { 19\lambda }{ 20 }$
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$\lambda$

A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio

$8:1$ . the ratio of de-broglie wavelengths of fragments are

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$1:2$
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$1:8$
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$4:1$
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None of these

Explanation

Momentum conversation. Given,

${ m }_{ 1 }{ V }_{ 1 }={ m }_{ 2 }{ V }_{ 2 }$

$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { 8 }{ 1 } =\dfrac { { m }_{ 2 } }{ { m }_{ 1 } }$

So,

$\dfrac { { m }_{ 1 } }{ { m }_{ 2 } } =\dfrac { 1 }{ 8 }$

but when particle break then density

$=$ constant

So,

$V\alpha m$

$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { { m }_{ 1 } }{ { m }_{ 2 } }$

$\dfrac { { a }_{ 1 }^{ 3 } }{ { a }_{ 2 }^{ 3 } } =\dfrac { 1 }{ 8 }$

$\dfrac { { a }_{ 1 } }{ { a }_{ 2 } } ={ \left( \dfrac { 1 }{ 8 } \right) }^{ 1/3 }\Rightarrow \boxed { \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { 1 }{ 2 } }$

$\Rightarrow { a }_{ 1 }:{ a }_{ 2 }=1:2$

If a strong diffraction peak is observed when electron are incident at an angle. 'i' from the normal to the crystal planes with distance 'd' wavelength them (see figure) de Broglie wavelength

$\lambda_{dB}$ of electrons can be calculate by the relationship (n is an integer)

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$d sin = n \lambda_{dB}$
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$2 d cos i= n \lambda_{dB}$
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$d sin = i \lambda_{dB}$
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$d cos = i \lambda_{dB}$

Explanation

Bragg's relation

$n\lambda =2d\sin\theta$ for having an in density maximum for diffraction pattern. But as the angle of incidence is given their

$n\lambda =2dcosi$ is the formula for finding a peat.

1232331_1312074_ans_85136ec3ade740ffa2c3d85512133c57.jpg

$1\mathop {\text{A}}\limits^{\text{0}}$

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$1.67 \times {10^{ - 27}}eV$
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$8.13 \times {10^{ - 2}}eV$
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$6.62 \times {10^{ - 22}}eV$
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$3.23 \times {10^{ - 2}}eV$

Explanation

Neutron

$m = 1.67\times 10^{-27}$ kg

$l = \frac{h}{p}$

$\Rightarrow P = \frac{h}{\lambda }$

$= \frac{66\times 10^{-34}}{1\times 10^{-10}}$ kg m/s

$\Rightarrow P = 6.6\times 10^{-24}$ kg m/s

energy

$E = \frac{P^{2}}{2m}= \frac{6.6\times 6.6\times 10^{-48}}{2\times 1.67\times 10^{-27}}$ joule

$= \frac{(6.6)^{2}}{3.34}\times 10^{-21}\times \frac{1}{1.6\times 10^{-19}}$ ev

E = 0.8 ev

1186092_1323773_ans_d8579a4d2fc94367923c5f0b40fe2cf8.jpg

The figure shows the path of white light's rays which leave in phase from two small source S

$_{1}$ and S

$_{2}$ and travel to a point X on a screen .The path difference is S

$_{2}$ X -- S

$_{1}$ X = 10

$\times 10^{-7}$ m.What wavelength of light give complete destructive interference at X?

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4.0 $\times 10^{-7}$ m
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6.6 $\times 10^{-7}$ m
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4.4 $\times 10^{-7}$ m
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5.8 $\times 10^{-7}$ m

If there is an increase in linear dimensions of the objects, the associated de-broglie wavelength.

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Increases
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Decreases
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Remains unchanged
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Depends on the density

Explanation

As we know,

By de Broglie wavelength equation,

Wavelength

$=$ plank's constant

$\div$ linear moment

i.e. wavelength

$\left( h \right) \alpha \dfrac { 1 }{ { \left( P \right) }_{ momentum } }$

If mass is increases momentum increases than de Broglie wavelength decreases.

If the deBroglie wavelenght of an electron is equal to

$10^{3}$ times the wavelength of a photon of frequency

$6 \times 10^{14} Hz$ , then the speed of electron is equal to : (Speed of light =

$3 \times 10^8 m/s$ Planck's constant =

$6.63 \times 10^{34} J$ . Mass of electron =

$9.1 10^{31} kg$ )

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$1.45\times 10^6m/s$
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$1.75\times 10^6m/s$
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$1.8\times 10^6m/s$
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$1.1\times 10^6 m/s$

Explanation

$\dfrac{h}{mv}=10^{-3}\left(\dfrac{3\times 10^8}{6\times 10^{14}}\right)$

$v=\dfrac{6.63\times 10^{-34}\times 6\times 10^{14}}{9.1\times 10^{-31}\times 3\times 10^5}$

If the particles listed below all have the same kinetic energy, which one would posses the shortest de-Broglie wavelength.

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Electron
0%
$\alpha -$ particle
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Proton
0%
All of the above

Explanation

As per the formula,

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

Since

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

thus,

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

Wavelength and mass has inverse dependence.

Since, mass of electron is the least out of all the three particles

$\therefore$ , electron has the largest de-Broglie wavelength.

If the de Broglie wavelengths associated with a proton and an

$\alpha$ -particle are equal then the ratio of velocities of the proton and the

$\alpha$ -particle will be:

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

Explanation

De - Broglie wavelength

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

wherein

$-h=$ plank's constant

$m=$ mass of particle,

$v=$ speed of particle

$E=$ Kinetic energy of particle

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

$\because$

${ \lambda }_{ \alpha }={ \lambda }_{ p }$

$\therefore$

${ p }_{ a }={ p }_{ b }$

$\therefore$

$m\alpha { v }_{ \alpha }={ m }_{ p }{ v }_{ p }$

$\therefore$

$\dfrac { { v }_{ p } }{ { v }_{ \alpha } } =\dfrac { { m }_{ a } }{ { m }_{ p } } =4:1$

What is difference between laser and normal light?

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Both light are polarized
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Normal light is polarized but laser light is not
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Both light are not polarized
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Normal light is not polarized but laser is polarized

A photon of energy

$8eV$ is incident on metal surface of threshold frequency

$1.6\times 10^{15}Hz$ . The maximum kinetic energy of the photoelectrons emitted (in eV) (Take

$h = 6\times 10^{-34} Js)$ .

0%
$1.4\ eV$
0%
$2.4\ eV$
0%
$4.8\ eV$
0%
$0.8\ eV$
0%
$7\ eV$

Explanation

As we know,

Work function

$W_0=hv_0=6.6\times 10^{-34}\times 1.6\times 10^{15}$

$=1.056\times 10^{-18}J=6.6\ eV$
From

$E=W_0+K_{max}\Rightarrow K_{max}=E-W_0 =1.4\ eV$

The de-Broglie wavelength

$(\lambda _B)$ associated with the electron orbiting in the second excited state of hydrogen atom is relared to that in the ground state

$(\lambda _G)$ by:

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$(\lambda _B=3 {\lambda _G})$
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$(\lambda _B=2 {\lambda _G})$
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$(\lambda _B=3 {\lambda _{G/3}})$
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$(\lambda _B=3 {\lambda _{G/2}})$

A non-monochromatic light is used in an experiment on the photoelectric effect. The stopping potential is related to the:

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Mean wavelength
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Longest wavelength
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Shortest wavelength
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None of the above

The graph between

$\dfrac { 1 } { \lambda }$ and stopping potential

$\mathrm { V } _ { 0 }$ of two metals having work functions

$\phi _ { 1 }$ and

$\phi _ { 2 }$ in an experiment of photoelectric effect is obtained as shown in the figure. Find out :

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Threshold wavelength of both metals
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$\phi _ { 1 } : \phi _ { 2 }$
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Which metal can emit photoelectrons with visible light?
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$\phi_{2 } : \phi_{2}$

A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency, then the number of photons emitted by the bulb in 20 seconds are

$\left( 1{ eV }=1.6\times { 10 }^{ -19 }J,hc=12400eV \right)$

0%
$2\times { 10 }^{ 18 }$
0%
${ 10 }^{ 18 }$
0%
${ 10 }^{ 21 }$
0%
$2{ \times 10 }^{ 21 }$

Explanation

$\begin{array}{l} Power\, of\, a\, bulb\left( p \right) =40W \\ energy\, emitted\, by\, a\, bulb=power\times Time\left( s \right) =40\times 20=800J \\ Energy\, of\, photons\, emitted\, by\, a\, bulb=\left( { \dfrac { { 80 } }{ { 100 } } } \right) \times 800=640J \\ Wavelength\left( \lambda \right) =620nm \\ Energy\, of\, a\, photon\, =\dfrac { { hc } }{ \lambda } =\dfrac { { 12400\times 1.6\times { { 10 }^{ -19 } }\times { { 10 }^{ -10 } } } }{ { 620\times { { 10 }^{ -9 } } } } \\ =\dfrac { { 640 } }{ { 3.2\times { { 10 }^{ -19 } } } } =2\times { 10^{ 21 } }\, photons \end{array}$

Hence,

option $(D)$ is correct answer.

A photon and an eletron both have wavelength 1A . The ratio of energy of photon to that of eletron is

0%
1
0%
0.012
0%
82.35
0%
$10^{-10}$

Explanation

Hence, option

$(C)$ is correct answer.
1380718_1358017_ans_83531ca446ed4655b44d8ab4832dee73.jpg

A photon falls through a height of

$1 km$ through the earth's gravitational field. To calculate the change in its frequency, take its mass to be

$hv/C^2$ . The fractional change in frequency

$v$ is close to :

0%
$10^{-20}$
0%
$10^{17}$
0%
$10^{-13}$
0%
$10^{10}$

Explanation

$hv'=hv+mgh$

$\therefore {h(v'-v)}{}=mgh$

$[\because mc^2=hv]$

$\therefore h(v'-v)=\dfrac{hv}{C^2}\times gh$

$\therefore \dfrac{v'-v}{v}=\dfrac{gh}{C^2}$

$=\dfrac{9.8\times 1\times 10^3}{(3\times 10^8)^2}$

$\therefore \dfrac{v'-v}{v}=1.1\times 10^{-13}$

If velocity of a particle is 3 times of that of electron and ratio of de brogile wavelength of particle to that of electron is

$1.814\times { 10 }^{ -4 }.$ The particle will be:-

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Neutron
0%
Deutron
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Alpha
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Tritium

Explanation

We know that,

$\dfrac { { m }_{ p } }{ { m }_{ e } } =\dfrac { 1.7\times { 10 }^{ -27 } }{ 9.1\times { 10 }^{ -31 } }$

$\dfrac { { m }_{ p } }{ { m }_{ e } }$ for alpha particle is -

Mass of

$1$ mole of

$\alpha$ particle

$=4g/mol$

Mass of

$1\alpha$ particle

$=\dfrac { 4 }{ 6.022\times { 10 }^{ -23 } } =6.64\times { 10 }^{ -27 }kg$

${ m }_{ p }/{ m }_{ e }$ for alpha particles

$=\dfrac { 6.64\times { 10 }^{ -24 } }{ 9.1\times { 10 }^{ -21 } } =7300$

${ m }_{ p }/{ m }_{ e }$ for deuterium particle

$=3650$

${ m }_{ p }/{ m }_{ e }$ for Tritium particle

$=\dfrac { 9.98\times { 10 }^{ -27 } }{ 9.1\times { 10 }^{ -31 } } =57.70$

$\therefore$

${ m }_{ p }/{ m }_{ e }$ ratio for neutron's is closest.

The energy of photon of visible light with maximum wavelength in

$eV$ is:

0%
$1$
0%
$1.6$
0%
$3.2$
0%
$7$

Explanation

Maximum wavelength of visible light (i.e. of red light) is

$7800\overset { 0 }{ A }$ .

Energy of red light

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

$E=\dfrac { 12400 }{ 7800 } V\overset { 0 }{ A }$

$\boxed { E=1.6eV }$

A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values

${ \lambda }_{ 1 }$ and

${ \lambda }_{ 2 }$ with

${ \lambda }_{ 1 }>{ \lambda }_{ 2 }.$ Which of the following statement is true?

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The particle could be moving in an circular orbit with origin as centre
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The particle could be moving in an parabolic orbit with origin as its focus
0%
When the de Broglie wave length is ${ \lambda }_{ 1 },$ the particle is nearer the origin than when its value is ${ \lambda }_{ 2 }$ .
0%
When the de Broglie wavelength is ${ \lambda }_{ 2 },$ the particle is nearer the origin than when its value is ${ \lambda }_{ 1 }$ .

Explanation

We know that,

${ \lambda }_{ 1 }=\dfrac { h }{ { mv }_{ 1 } }$ and

${ \lambda }_{ 2 }=\dfrac { h }{ { mv }_{ 2 } }$

$\therefore$

$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 1 } } =\dfrac { { v }_{ 2 } }{ { v }_{ 1 } }$ and thus

${ \lambda }_{ 1 }>{ \lambda }_{ 2 }$

By lower of conversation of angular momentum the particle moves faster when it is closer to focus.

The de Broglie wavelength of an electron moving with a velocity

$2.25 \times 10^8$

$ms^{-1}$ is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of photon is :

0%
$\dfrac{3}{8}$
0%
$4$
0%
$\dfrac{1}{2}$
0%
$2-\sqrt 3$

Explanation

$K.E_{particle}=\dfrac 12 mv^2$

Also,

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

$\implies K.E_{particle}=\dfrac 12(\dfrac{h}{\lambda v})v^2=\dfrac{vh}{2\lambda}$ ............(i)

$K.E_{photon}=\dfrac{hc}{\lambda}$ ..............(ii)

Therefore,

$\dfrac{K.E_{particle}}{K.E_{photon}}=\dfrac{v}{2c}=\dfrac{2.25\times 10^8}{2\times 3\times 10^8}=\dfrac 38$

During an experiment an

$\alpha$ -particle and a proton are accelerated by same positive difference , their de broglie wavelength ratio will (Take mass of proton= mass of neutron).

0%
$1:2$
0%
$1:4$
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$1:2\sqrt { 2 }$
0%
$1:\sqrt { 2 }$

You are given an electron with a deBroglie wavelength of

$\lambda =76.3$ nm. What is the Kelvin temperature of this electron ?

0%
1.50
0%
2.00
0%
2.50
0%
3.00

Explanation

$\begin{array}{l} \lambda =\dfrac { h }{ p } \, \, and\, \, \, \dfrac { { { p^{ 2 } } } }{ { 2m } } =\varepsilon =\dfrac { 3 }{ 2 } kT \\ \therefore p={ \left( { 3mkT } \right) ^{ \dfrac { 1 }{ 2 } } } \\ \lambda =\dfrac { h }{ p } =\dfrac { h }{ { { { \left( { 3kmT } \right) }^{ \dfrac { 1 }{ 2 } } } } } \\ \therefore T=\dfrac { { { h^{ 2 } } } }{ { 3km{ \lambda ^{ 2 } } } } \\ T=\dfrac { { { h^{ 2 } } } }{ { 3k{ \lambda ^{ 2 } } } } =\dfrac { { { { \left( { 6.626\times { { 10 }^{ -34 } } } \right) }^{ 2 } } } }{ { \left( 3 \right) \left( { 1.38\times { { 10 }^{ -23 } } } \right) \left( { 9.108\times { { 10 }^{ -31 } } } \right) { { \left( { 76.3\times { { 10 }^{ -9 } } } \right) }^{ 2 } } } } =2.00K \\ T=2.00K \end{array}$

Option $B$ is correct .

During an experiment an

$\alpha$ - particle and a proton are having a difference , their de Broglie wavelenght ratio will be

0%
$1:2$
0%
$1:4$
0%
$1:2\sqrt { 2 }$
0%
none of these

Explanation

Hence, option

$(C)$ is correct answer.
1380758_1389172_ans_632b2ccd5be641ae9122bdf2b67a82fd.jpg

A photon collides with a stationary hydrogen atom is ground state inelastically. Energy the colliding photon is

$10.2 eV$ . After atime interval of the order of micro second another photon collides with same hydrogen atom inelatically with an energy of

$15 eV$ . What will be observed by the detector?

0%
$2$ photons of energy $10.2 eV$
0%
$2$ Photons of energy $1.4 eV$
0%
One photon of energy $10.2 eV$ and an electron of energy $1.4 eV$
0%
One photon of energy $10.2 eV$ and another photon of energy $1.4 eV$

Explanation

According to the question,

Due to

$10.2\ eV$ photon one photon of energy

$10.2\ eV$ will be detected.
Due to

$15\ eV$ photon the electron will come out of the atom with energy

$(15-13.6)=1.4\ eV$ .

If the de-Broglie wavelengths associated with a proton and

$\alpha$ -particle are equal, then the ratio of velocities of the proton and the a-particle will be:

0%
$4:1$
0%
$2:1$
0%
$1:2$
0%
$1:4$

Explanation

$\begin{array}{l} { \lambda _{ p } }=\dfrac { h }{ { m{ V_{ 1 } } } } \\ { \lambda _{ \alpha } }=\dfrac { h }{ { 4m{ v_{ 2 } } } } \\ \Rightarrow \dfrac { h }{ { m{ v_{ 1 } } } } =\dfrac { h }{ { 4m{ v_{ 2 } } } } \\ \Rightarrow { v_{ 1 } }=4{ v_{ 2 } } \\ \Rightarrow \dfrac { { { v_{ 1 } } } }{ { { v_{ 2 } } } } =4:1 \\ Hence, \\ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

An electron with an initial kinetic energy of

$100\ eV$ is accelerated through a potential difference of

$50\ V$ . Now the de-Broglie wavelength of electron becomes

0%
$1\ \overset{o}{A}$
0%
$\sqrt{1.5}\ \overset{o}{A}$
0%
$\sqrt 3\ \overset{o}{A}$
0%
$12.27\ \overset{o}{A}$

Explanation

The initial kinetic energy of the electron is

$100eV$

Now, it is accelerated through a potential is

$50V$

Therefore, the additional kinetic energy it gains is

$K'=eV=50\,eV$

Hence, the total kinetic energy possessed by the electron is

$150\,eV$ which is equivalent to

$150\times 1.6\times 10^{-19}\,J$

$\lambda =\dfrac{h}{\sqrt{2mK}}=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 150\times 1.6\times 10^{-19}}}=1\times 10^{-10}\,m=1\ \overset{o}{A}$

The smallest quantum of energy is:-

0%
All
0%
Electron
0%
Proton
0%
Photon

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature

$27^oC$ and

$127^oC$ respectively is

0%
$\dfrac{1}{2}$
0%
$2$
0%
$\sqrt{\dfrac{8}{3}}$
0%
$1$

If h is the planck's constant, the momentum of a photon of wavelength 0.01 A is :

0%
$h \times 10^{-2}$
0%
h
0%
$h \times 10^2$
0%
$h \times 10^{12}$

Explanation

By de-Broglie formula

$\lambda=\dfrac hp$

So,

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

$=\dfrac{h}{0.01\times10^{-10}}=h\times 10^{12}$

If a photon having wavelength

$6.2 { nm }$ was allowed to strike a metal plate having work function

$50 { eV }$ , then calculate wavelength associated with emitted electron.

0%
$1 \times 10 ^ { - 10 } \mathrm { m }$
0%
$2\times 10 ^ { - 10 } \mathrm { m }$
0%
$3 \times 10 ^ { - 18 } \mathrm { m }$
0%
$4 \times 10 ^ { - 14 } \mathrm { m }$

Explanation

Energy of photon

$=\dfrac{12400}{\lambda(in\,A^0)}eV$

$=\dfrac{12400}{62}$

$=200eV$

$KE_{max}=Energy\,\,.of \,\,photon-work function$

$=200-50$

$=150eV$

We know,

$\lambda=\dfrac{h}{\sqrt{2M(KE_{max})}}$

$=\dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times (150\times 1.6\times 10^{-19})}}$

$=\dfrac{6.6\times 10^{-34}}{66\times 10^{-25}}$

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

$=1A^0$

A photon of wavelength

$\lambda$ is incident on a free electron at rest and scattered in backward direction with a wavelength

$2\lambda$ . The ratio of kinetic energy and movement of the electron will be (Assume Newtonian mechanics applies)

0%
$C 3$
0%
$C 6$
0%
$C 9$
0%
$C {12}$

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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