MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 4

The de Broglie wavelength of an electron which falls through a p.d. of 10,000V is

0%
1.227 $A^{0}$
0%
12.27 $A^{0}$
0%
0.1227 $A^{0}$
0%
2.454 $A^{0}$

Explanation

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

$={\dfrac{6.62\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 10^4}}}$

$= 0.1227 A^o$

So, the answer is option (C).

A proton has kinetic energy

$\mathrm{E}=100\mathrm{k}\mathrm{e}\mathrm{V}$ which is equal to that of a photon. The wavelength of photon is

$\lambda_{2}$ and that of proton is

$\lambda_{1}$ . The ratio of

$\lambda_{1}/\lambda_{2}$ is proportional to

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

Explanation

For proton and photon wavelength is given as -

$\displaystyle \lambda_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}}=\frac{\mathrm{h}}{\sqrt{2\mathrm{m}\mathrm{E}}},\ \displaystyle \lambda_{\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}}=\frac{\mathrm{h}\mathrm{c}}{\mathrm{E}}$ .

So, the answer is option (B).

The wavelength

$\lambda_e$ of an electron and

$\lambda_p$ of a photon of same energy E are related by -

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$\lambda_p$ $\lambda_e^2$
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$\lambda_p$ $\lambda_e$
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$\lambda_p$ $\sqrt{\lambda_e}$
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$\lambda_p$ $\frac{1}{\sqrt{\lambda_e}}$

Explanation

Wavelength of electron

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

and proton

$={ \lambda }_{ p }=\dfrac { hC }{ E }$

$\therefore$

${ \lambda }_{ p }\alpha { \lambda }_{ e }^{ 2 }\Rightarrow { \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2mE }$ or

$E=\dfrac { hC }{ { \lambda }_{ p } }$

$\therefore$

${ \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2m\dfrac { hc }{ { \lambda }_{ p } } }$

$\Rightarrow { \lambda }_{ e }^{ 2 }=\dfrac { { h }^{ 2 } }{ 2mhc } { \lambda }_{ p }$

$\boxed { \therefore \quad { \lambda }_{ e }^{ 2 }\alpha { \lambda }_{ p } }$

The hydrogen atom emits a photon of 656.3nm line. The momentum of the photon associated with it is

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$10^{-27}kgms^{-1}$
0%
$10^{-23}kgms^{-1}$
0%
$10^{-18}kgms^{-1}$
0%
$10^{-15}kgms^{-1}$

Explanation

$\lambda = 656.3 nm$

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

$= \dfrac{6.6\times 10^{-34}}{656.3\times 10^{-9}}$

$={10^{-27} kg ms^{-1}}$

So, the answer is option (A).

The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement ?

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Curves (b) and (c) represent incident radiations same frequencies having same

intensity.
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Curves (a) and (b) represent incident radiations of different frequencies and

different intensities
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Curves (a) and (b) represent incident radiations of same frequencies but of

different intensities
0%
Curves (b) and (c) represent incident radiations of different frequencies and

different intensities

Monochromatic light of frequency

$6.0\times 10^{14}$ Hz is produced by a laser. The power emitted is

$2\times 10^{-3}$ W. The number of photons emitted, on the average, by the source per second is:

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$5\times 10^{14}$
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$5\times 10^{15}$
0%
$5\times 10^{16}$
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$5\times 10^{17}$

Explanation

No. of photoelectrons emitted,

$n=\dfrac{p}{hv}.=\dfrac{2\times 10^{-3}}{6.62\times 10^{-34}\times 6\times 10^{14}}=5\times 10^{15}$

The de Broglie wavelength of a neutron when its kinetic energy is K is

$\lambda$ .What will be its wavelength when its kinetic energy is 4K ?

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

Explanation

if KE has become 4 times, then speed must have doubled.

and

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

i.e.

$\lambda \alpha \dfrac{1}{V}$

${\lambda}_1 V_1 = {\lambda}_2 V_2$

$\lambda V = \lambda (2V)$

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

So, the answer is option (B).

The de Broglie wavelength associated with an electron of velocity 0.3c and rest mass

$9.1 \times 10^{-31}$ kg is

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$7.68\times 10^{-10}m$
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$7.68\times 10^{-12}m$
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$5.7\times 10^{-12}m$
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$9.1\times 10^{-12}m$

Explanation

The de-broglie wavelength according to formula

$=\dfrac{h}{m_ov}\sqrt{(1-\dfrac{v^2}{c^2})}$

$\Rightarrow \lambda =\dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times0.3\times3\times10^{8}}\sqrt{1-(\dfrac{0.3c}{c})^2}$

$\Rightarrow \lambda =7.68\times10^{-12}m.$

So, the answer is option (B).

A source S

$_{1}$ is producing 10

$^{15}$ photons per second of wavelength 5000

$\dot{a}$ . Another sorce S

$_{2}$ is producing

$1.02 \times 10^{15}$ photons per second of wavelength 5100

$\dot{A}$ .Then, (power of S

$_{2}$ )/(power of S

$_{1}$ ) is equal to

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1.00
0%
1.02
0%
1.04
0%
0.98

Explanation

Power

$S_1=\dfrac{n_1\times hc}{\lambda_1 \times 1 second}$

Power

$S_2=\dfrac{n_2\times hc}{\lambda_2 \times 1 second}$

the ratio

$\dfrac{S_2}{S_1}=\dfrac{n_2\times \lambda_1}{n_1\times \lambda_2}=\dfrac{1.02\times 10^{15}\times 5000}{10^{15}\times 5100 }=1.00$

So the correct answer is 1.00

The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is:

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$4125 A^o$
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$3000 A^o$
0%
$6000 A^o$
0%
$2062 A^o$

Explanation

Since work function for a metal surface is

$W=\dfrac {hc}{\lambda_0}$
Where

$\lambda_0$ is threshold wavelength or cut-off wavelength here

$W=4.125 eV=4.125\times 1.6\times 10^{-19} Joule$

$\lambda_0=\dfrac {6.6\times 10^{-34}\times 3\times 10^8}{4.125\times 1.6\times 10^{-19}}=3000 A^o$

A beam of light of wavelength

$\lambda$ is incident on a mirror at an angle

$\theta$ . Find the change in momentum of the photon after reflection.

0%
$2 \dfrac{h}{\lambda}sin \theta$
0%
$\dfrac{h}{\lambda} cos \theta$
0%
$2 \dfrac{h}{\lambda}cos \theta$
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none of these

Explanation

After reflection the horizontal component of velocity remains same and normal component reverses its direction

$\Delta P = 2 P\cos \theta$

For photon

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

$\Delta P = 2 \dfrac{h}{\lambda}\cos \theta$

Mass of moving photon is

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$\frac {hv}{c^2}$
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$\frac {hv}{c}$
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hv
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Zero

The momentum of a photon having energy E is

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$\frac {E}{c}$
0%
$\frac {E}{c^2}$
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$\frac {E}{h}$
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Zero

A Photoelectric cell converts

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Light energy into electrical eneregy
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Light energy into sound energy
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Electric energy into light energy
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Light energy into heat energy

When light is incident on a surface, photoelectrons are emitted.

For those photoelectrons:

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The value of kinetic energy is same
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Kinetic energy does not depend on the wave length of incident light
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The value of kinetic energy is equal to or less than a maximum energy
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None of the above

Explanation

Kinetic energy of photo electron is

$KE=h\nu -W=\frac { hc }{ \lambda } -W\\ h=plancks\quad constant\\ \nu =frequency\\ \lambda =wavelength\quad of\quad incident\quad light\quad \\ W=work\quad force\\$

The photo electrons are not emitted instantly. They collide several times with other electrons within the metal before coming out. Hence the value of kinetic energy is not the same. From above equation, Kinetic energy depends on wavelength of incident light.Also the value of kinetic energy is less than or equal to

$h\nu$

In a photoelectric effect, the anode potential is plotted against the plate current

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A and B will have different intensities while B and C will have different frequencies
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B and C will have different intensities while A and C will have different frequencies
0%
A and B will have different intensities while A and C will have equal frequencies
0%
A and B will have equal intensities while B and C will have different frequencies

Rest mass of a photon is

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$\frac {hv}{c^2}$
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$\frac {hv}{c}$
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$hv$
0%
zero

Explanation

$\textbf{Hint}$ : Use relativistic mass formula.

$\textbf{Step1:Photon}$

According to Einstein's quantum theory light propagates in the form of packets i.e. quanta of energy, which is called photon.

$\textbf{Step2:Rest mass of photon}$

The rest mass of photons is being zero. It can be shown, according to the relativity theory of light.

According to the relativistic theory equation, the mass of the photon is computed as:

$m = \dfrac{m_{0}}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}$

$m_0=m {\sqrt{1-\dfrac{v^2}{c^2}}}$

when $v=0$ ;

So $m_0=0$

Where $m_0$ is the rest mass of the Photon.

The photoelectric effect is described as the ejection of electrons from the surface of a metal when

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It is heated to a high temperature
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Light of suitable frequency falls on it
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Electrons of suitable velocity impinge on it
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It is placed in a strong magnetic field

Which of the following systems may be adequately described by classical physics?

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Motion of a cricket ball
0%
Motion of dust particle
0%
A hydrogen atom
0%
A neutron changing to a proton

Explanation

The formulation or rules of classical physics are quite accurate for heavy bodies having finite velocities like the planets, vehicles etc. For particle much smaller than

$10^{-6} m$ (such as atoms, nuclei etc.) these rules do not work well.

A point of source of 6 watts emits monochromatic light of wavelength

$5000\dot A$ . The number of photons striking normally per second per unit area of the surface distant 5 m from the source will be

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$4.82$
0%
$4.82 \space \times \space 10^{-4}$
0%
$4.82 \space \times \space 10^{-6}$
0%
$4.82 \space \times \space 10^{16}$

Explanation

The photons will be emitted symmetrically in all directions.

Power of Source =

$6W$

$Energy=nh\nu=\dfrac{nhc}{\lambda}$ where

$n$ is no.of

$photons$ emitted per second.

Soving the above equation we get n.

The no.of photons falling on

$1m^2$ area is equal to

$\dfrac{n}{4\pi d^2}$ where

$d$ is the distance of the point.

$\Rightarrow$ No.of photons=

$\dfrac{1.51\times10^{19}}{4\pi(5)^2}=4.82\times 10^{16}$

So, the answer is option (D).

A 40 watt bulb converts 6% of its power to red light (wavelength

$6500 \dot{A}$ ). The number of red lights photons emitted by the bulb per second is

0%
$100$
0%
$4 \times 10^{18}$
0%
$8 \times 10^{18}$
0%
$13 \times 10^{18}$

Explanation

In 1 second the bulb would consume 40J of energy, out of which only 6% is radiated as red light
In other words only 2.4J of energy is red light.

Each photon of red light has (E) =

$\dfrac{hc}{\lambda}$ ; where h is plancks constant, and c is the speed of light.

Number of photons is equal to the total energy divided by the energy of each photon =

$\dfrac{2.4\lambda}{hc}$

$\dfrac{2.4*6500*10^{-10}}{6.64*10^{-34}*3*10^{8}}$ =

$8 * 10^{18}$

So, the answer is option (C).

A radio station emits

$10kW$ power of

$90.8 MHz$ . Find the number of photons emitted per second

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$1.6 \times \displaystyle 10^{28}$
0%
$\displaystyle 1.6\times 10^{29}$
0%
$\displaystyle 1.6\times 10^{30}$
0%
$\displaystyle 1.6\times 10^{32}$

Explanation

We have, number of the photons emitted per sec,

$N = \dfrac {Power}{Energy \ of \ the \ Photon} = \dfrac {P}{h \nu}$

$\Rightarrow N = \dfrac{ 10 \times 10^3} {6.626 \times 10^{-34} \times 90.8 \times 10^6}$

$\Rightarrow N = 1.6 \times 10^{29}$ photons per sec

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

$652 \displaystyle nm$ in pulses that are of

$20 \displaystyle ms$ duration. The average power during each pulse is

$0.6 \displaystyle W$ . Find the energy in each pulse in

$\displaystyle eV$ and in a single photon.

0%
$\displaystyle 7.5\times 10^{15} eV, 1.9 eV$
0%
$\displaystyle 7.5\times 10^{15} eV, 1.19 eV$
0%
$\displaystyle 7.5\times 10^{16} eV, 0.19 eV$
0%
$\displaystyle 7.5\times 10^{16} eV, 1.9 eV$

Explanation

Energy of the 652 nm photon is:

$\displaystyle E(eV)=\dfrac{1240}{652}=1.9eV$ (in a single photon)

The pulse duration is 20 ms, so,

Energy in 20

$\displaystyle ms = 0.6\times20\times10^{-3}=1.2\times10^{-2}J$

$\displaystyle=\dfrac{1.2\times10^{-2}} {1.6\times10^{-19}}eV$

$\displaystyle=7.5\times10^{16}eV$

So, the answer is option (D).

A TV center transmits 10 kilowatt of power at 150 MHz. The energy of a photon of electromagnetic wave is

0%
$6\space \times 10^{-7} \space J$
0%
$6\space \times 10^{-7} \space eV$
0%
$6\space \times 10^{-17} \space J$
0%
$6\space \times 10^{-17} \space eV$

Explanation

The energy of a photon of electromagnetic wave is

$E=h\nu$
where,

$h$ is Planck's constant,

$\nu$ is frequency of photon
then

$E=6.626\times { 10 }^{ -34 }\times 150\times { 10 }^{ 6 }\quad J\\ \quad =9.94\times { 10 }^{ -26 }\quad J\\ \quad =\dfrac { 9.94\times { 10 }^{ -26 } }{ 1.6\times { 10 }^{ -19 } } \quad eV\\ \quad =6.21\times { 10 }^{ -7 }\quad eV$
So, the answer is option (B).

The energy of a photon of wavelength

$6000\dot A$ in eV will be

0%
1.06 eV
0%
0.206 eV
0%
2.06 eV
0%
20 eV

Explanation

As we know that amount of energy emitted in electron volt is given as,

${E}_{ev}=\dfrac{12420}{\lambda}$

So,

${E}=\dfrac{12420}{6000}={2.06} \ electron \ volt$

So, the answer is option (C).

A. The energy

$E$ and momentum

$p$ of a photon are related as

$p=\dfrac{E}{c}$ .

R. The photon behaves as a particle.

0%
$A$ and $R$ are both correct and $R$ is correct explanation of $A$
0%
$A$ and $R$ are correct but $R$ is not correct explanation of $A$
0%
$A$ is correct but $R$ is false
0%
Both $A$ and $R$ are false

Explanation

Assertion is correct and reason is too correct but it is not a correct explanation.
Suppose if a particle as kinetic energy

$K$ then its velocity will be

$\sqrt{\dfrac{K}{2m}}$ where m is mass of particle , then its momentum is mass times its velocity .

So photon is a particle will just not be sufficient to explain the assertion.

So, the answer is option (B).

If a proton and an electron have the same de Broglie wavelength, the ratio of the velocity of proton to the velocity of electron will be nearly

0%
$1$
0%
$1840$
0%
$\displaystyle\dfrac{1}{1840}$
0%
$44$

Explanation

De Broglie wavelength is given by:

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

If two bodies (electron and proton) have the same De Broglie wavelength, they have the same momentum.
i.e.

$p_{1} = p_{2}$

$m_{e}v_{e} = m_{p}v_{p}$
i.e.

$\dfrac{m_{e}}{m_{p}} = \dfrac{v_{p}}{v_{e}} = \dfrac{1}{1840}$

So, the answer is option (C).

The hydrogen atom emits a photon of 656.3 nm line. Find the momentum of the photon associated with it.

0%
$10^{-27}$ kg $ms^{-1}$
0%
$10^{-23}$ kg $ms^{-1}$
0%
$10^{-25}$ kg $ms^{-1}$
0%
none of these

Explanation

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

$=\dfrac{6.625 \times10^{-34}}{656.3 \times 10^{-9}}$

$= 1.0 \times 10^{-27} kg ms^{-1}$

So, the answer is option (A).

The energy that should be added to an electron to reduce its de broglie wavelength from 1 nm to 0.5 nm is

0%
Four times the initial energy.
0%
Equal to initial energy.
0%
Twice the initial energy.
0%
Thrice the initial energy.

Explanation

$\ \lambda =\dfrac{h}{p}=\dfrac{h}{\sqrt{2m(KE)}}$

$\therefore$ New energy should be 4 times and hence energy to be added is thrice the initial energy.

So, the answer is option (D).

The wavelength associated with 1

$MeV$ proton is

0%
28.6 $pm$
0%
2.86 $pm$
0%
2.86 $fm$
0%
28.6 $fm$

Explanation

The wavelength and the energy is related by:

$\displaystyle \lambda =\dfrac{0.286}{\sqrt{10^{6}}}=28.6fm$

So, the answer is option (D).

The minimum energy required to dissociate

$Ag$

$Br$ bond in 0.6

$eV$ . A photographic flim is coated with a sliver bromide layer. Find the maximum wavelength whose signature can be recorded on the film

0%
207 $nm$
0%
702 $nm$
0%
207 $A^{\circ}$
0%
2070 $nm$

Explanation

We have

$E = \dfrac {1240}{\lambda} eV$

$\implies \lambda = \dfrac {1240}{E} = \dfrac {1240}{0.6} = 2070 nm$

So, the answer is option (D).

Find photon energy in

$eV$

0%
1.04 $eV$
0%
1.14 $eV$
0%
1.72 $eV$
0%
1.28 $eV$

Explanation

The energy of the incident photon that is just able to dissociate an AgBr molecule is,

$E =\dfrac{10^{5}}{6.023\times 10^{23}\times 1.6\times 10^{19}}=1.04eV$

So, the answer is option (A).

Find the wavelength of 100

$eV$ electron

0%
$1.227 \displaystyle A^{\circ}$
0%
$1.72 \displaystyle A^{\circ}$
0%
$1.24\displaystyle nm$
0%
$12.4 \displaystyle nm$

Explanation

We have

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

$\implies \lambda = \dfrac {6.625 \times 10^{-34}}{\sqrt {2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}} = 1.227 A^o$

So, the answer is option (A).

A surface has work function 3.3

$eV$ . Which of the following will cause emission?

0%
100 $W$ incandascent lamp
0%
40 $W$ flouroscent lamp
0%
20 $W$ sodium lamp
0%
20 $W$ Hg lamp

Explanation

Since

$\lambda_{min} =400$ is the minimum wavelength of visible region we will use it in the expression

$E = \dfrac {1240}{\lambda} eV$

$\implies E = \dfrac{1242}{400} =3.1 eV.$

$\therefore$ In order for emission we require a source of uv light, since visible light can not cause emission. In the given options, only Hg lamp is capable of generating uv light.

So, the answer is option (D).

Find the wavelength of 10

$MeV \ \alpha-$ particles

0%
3 $A^{\circ}$
0%
3 $pm$
0%
3 $fm$
0%
30 $fm$

Explanation

The wavelength and the energy is related by:

$\displaystyle \lambda (A^{\circ})=\frac{0.286}{\sqrt{10^{7}}}=0.3\times10^{-3}A^{\circ}=3\times10^{-15}m=3fm$

So, the answer is option (C).

In the following diagram if

$\displaystyle V_{2}>V_{1}$ then

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$\displaystyle \lambda _{1}=\sqrt{\lambda _{2}}$
0%
$\displaystyle \lambda _{1}<\lambda _{2}$
0%
$\displaystyle \lambda _{1}=\lambda _{2}$
0%
$\displaystyle \lambda _{1}>\lambda _{2}$

Explanation

Using Einstein's equation of photoelectric effect:

$eV_1=\dfrac{hc}{\lambda_2}-W$ ...(1)

where,

$W=$ work function of metal

and

$eV_2=\dfrac{hc}{\lambda}_2-W$ ...(2)

From (1) and (2):

$eV_2-eV_1=\dfrac{hc}{\lambda_2}-\dfrac{hc}{\lambda_1}$

$V_2-V_1=\dfrac{hc}{e}(\dfrac{1}{\lambda_2}-\dfrac{1}{\lambda_1})$

$V_2>V_1$ so

$(1/\lambda_2-1/\lambda_1)>0$ or

$1/\lambda_2>1/\lambda_1$ or

$\lambda_1>\lambda_2$

So, the answer is option (D).

The work function of a metal is 2.5 eV. When photon of some proper energy is made incident on it, then an electron of 1.5 eV is emitted. The energy of photon will be

0%
$4 eV$
0%
$1 eV$
0%
$1.5 eV$
0%
$2.5 eV$

Explanation

The energy of photoelectron emitted out is

$E = h\nu - \phi$
where,

$h\nu$ is the energy of incident photon,

$\phi$ is the photoelectric work function of the metal.

$h\nu = E + \phi$

$h\nu = 1.5 + 2.5$

$h\nu = 4 eV$
Hence, the energy of photon will be 4eV.

So, the answer is option (A).

The number of photons of wavelength

$13.2$

$A^{0}$ in

$6$

$J$ of energy is:

(

$h$ =

$6.6\times10^{-34}$

$Js$ )

0%
$2\times10^{12}$
0%
$4\times10^{16}$
0%
$6\times10^{20}$
0%
$4\times10^{24}$

Explanation

The energy of one photon of wavelength

$13.2 A$ is

$E = \dfrac{hc}{\lambda} = \dfrac{6.64 \times 10^{-34} \times 3 \times 10^{8}}{13.2 \times 10^{-10}} = 1.509 \times 10^{-16} J$

Hence 6J of energy must contain

$\dfrac{6}{1.509 \times 10^{-16}}$ photons

$= 4 \times 10^{16}$ photons

So, the answer is option (B).

The relation between energy E and momentum p of a photon is

0%
$E = pc$
0%
$E=\dfrac { p }{ c }$
0%
$p = Ec$
0%
$E=\dfrac { { p }^{ 2 } }{ c }$

Explanation

The energy of a photon in vacuum is equal to

$E = pc$ (where p is the momentum of the photon, and c is the speed of light in vacuum)

So, the answer is option (A).

The effective mass of photon of wavelength 40

$\overset { \circ }{ A }$ will be:

0%
$55.2\times { 10 }^{ -35 }kg$
0%
$55.2\times { 10 }^{ -33 }gm$
0%
$55.2\times { 10 }^{ -17 }kg$
0%
$55.2\times { 10 }^{ -38 }kg$

Explanation

From de Broglie hypothesis, wavelength of photon ,

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

$h=$ Planck's constant ,

$m=$ effective mass and

$c=$ velocity of light or photon.

So,

$m=\dfrac{h}{\lambda c}=\dfrac{6.62\times 10^{-34}}{(40\times 10^{-10})\times (3\times 10^8)}=55.2\times 10^{-35} kg$

Two electrons are moving with the same velocity, one electron enters a region of uniform electric field while the other enters a region of uniform magnetic field. Then, after sometime, if the de-Broglie wavelength of two are

$\lambda_{1}$ and

$\lambda_{2}$ respectively, then

0%
$\lambda_{1} = \lambda_{2}$
0%
$\lambda_{1} > \lambda_{2}$
0%
$\lambda_{1} < \lambda_{2}$
0%
$\lambda_{1}>\lambda_{2}$ or $\lambda_{1}< \lambda_{2}$

Explanation

Energy of an electron

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

where,

$\lambda$ is the de-Broglie wavelength of the electron.

Therefore, Energy is inversely proportional to the de-Broglie wavelength of an electron.

After some time, electron that has entered electric field has an increase in velocity along the direction of field, which results in the increase of translational kinetic energy without saturation.

The electron which entered the magnetic field has an increased rotational kinetic energy.

Hence, the electron in electric field will have more energy and low de-Broglie wavelenth.

$\Rightarrow \lambda_2>\lambda_1$

So, the answer is option (C).

Find the minimum wavelength of X-ray produced if 10 kV potential difference is applied across the anode and cathode of the tube.

0%
12.4 A $^{\circ}$
0%
12.4 nm
0%
1.24 nm
0%
1.24 A $^{\circ}$

Explanation

If electrons are accelerated to a velocity

$v$ by a potential difference V and then allowed to collide with a metal target, the maximum frequency of the X-rays emitted is given by the equation:

$mv^2= eV = h\nu$

or,

$\displaystyle \lambda_{min} = \dfrac{1240}{10^3}*10^{-9} = 124 nm = 12.4 A^{\circ}$

So, the answer is option (D).

The correct curve between the energy of photon (E) and its wave length

$\ \lambda$ is

Explanation

The energy equation for a photon in terms of wavelength

$(\lambda)$ is given by,

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

$h=$ Plank's constant and

$c=$ velocity of photon.

So its like a

$y=1/x$ graph. Hence the option A will represent the correct graph.

So, the answer is option (A).

A radio transmitter is working at frequency 880 kHz and power 10kW. The number of photons emitted per second will be

0%
$\displaystyle 1327\times 10^{34}$
0%
$\displaystyle 0.075\times 10^{-34}$
0%
$\displaystyle 1.71\times 10^{31}$
0%
$\displaystyle 13.27\times 10^{34}$

Explanation

The energy of photon is given by

$E = h\nu$
where h is Plank's constant,

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

$\nu$ is the frequency of the photon.

$E = 6.6\times 10^{-34} \times 880 \times 10^3$

$E = 5808 \times 10^{-31}$

$E = 5.808 \times 10^{-28} J$
Now, for the n number of photons emitted per second we can write

$nE = P$ ................................(since, power is energy (emitted or absorbed) per second)

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

$n = \dfrac{10 \times 10^3}{5.808 \times 10^{-28}}$

$n = 1.721 \times 10^{31}$

So, the answer is option (C).

The momentum of photon of frequency

${ 10 }^{ 14 }Hz$ will be

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$2.2\times { 10 }^{ -26 }Kg \ m/sec$
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$2.21\times { 10 }^{ -28 }Kg \ m/sec$
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${ 10 }^{ -28 }Kg \ m/sec$
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$0.21\times { 10 }^{ -2 }Kg \ m/sec$

Explanation

From de Broglie hypothesis, wavelength of photon ,

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

$\dfrac{c}{f}=\dfrac{h}{p}$ where

$h=$ Planck's constant ,

$p=$ momentum ,

$f=$ frequency and

$c=$ velocity of light or photon.

So,

$p=\dfrac{hf}{ c}=\dfrac{(6.62\times 10^{-34})\times 10^{14}}{ (3\times 10^8)}=2.21\times 10^{-28} Kg \ m/sec$

So, the answer is option (B).

Protons are accelerated from rest by a potential difference 4 kV and strike a metal target. If a proton produces one photon on impact of minimum wavelength

$\lambda_1$ and similarly an electron accelerated to 4 kV strikes the target and produces a minimum wavelength

$\lambda_2$ then

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$\lambda_1 = \lambda_2$
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$\lambda_1 > \lambda_2$
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$\lambda_1 < \lambda_2$
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no such relation can be established

Explanation

If electrons are accelerated to a velocity

$v$ by a potential difference V and then allowed to collide with a metal target, the maximum frequency of the X-rays emitted is given by the equation:

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

or,

$\displaystyle \lambda_{min} = \dfrac{1240}{V}$ .

So this shows that the minimum wavelength is inversely proportional to the accelerating voltage. Here accelerating voltage are same so the minimum wavelength is same.

So, the answer is option (A).

Neglecting variation of mass with energy the wavelength associated with an electron having a kinetic energy

$E$ is proportional to

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

Explanation

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

momentum p = mv

this gives

$p^2=2Em$

also

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

When an electron moving at a high speed strikes a metal surface, which of the following are possible?

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The entire energy of the electron may be converted into an X-ray photon
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Any fraction of the energy of the electron may be converted into an X-ray photon.
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The entire energy of the electron may get converted to heat.
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The electron may undergo elastic collision with the metal surface.

An electron makes transition from

$n=4$ to

$n=1$ state in a hydrogen atom. The maximum possible number of photons emitted will be :

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$1$
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$2$
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$3$
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$6$

Explanation

From the nth state, the electron may go to

$(n - 1)^{th}$ state, ...., 2nd state or 1st state. So there are (n 1) possible transitions starting from the

$n^{th}$ state. The atoms reaching

$(n - 1)^{th}$ state may make (n 2) different transitions. The atoms reaching

$(n - 2)^{th}$ state may make (n 3) different transitions. Similarly for other lower states. During each transition a photon with energy

$h\nu$ and wavelength

$\lambda$ is emitted out. Hence, the total number of possible transitions is equal to the number of photons emitted. The total number of possible transitions is

$(n-1), (n-2), (n-3), ................, 3, 2, 1 = \dfrac{n(n-1)}{2}$

Therefore, for transition of an electron from higher energy state n = 4 to lower energy state n = 1 the number of photons emitted are

$\dfrac{n(n-1)}{2} = \dfrac{4(4-1)}{2} = \dfrac{12}{2} = 6$

On increasing the applied potential difference in X-ray tube

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The intensity of emitted radiation increases.
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The minimum wavelength of emitted radiation increases.
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The intensity of emitted radiation remains unchanged.
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The minimum wavelength of emitted radiation decreases.

Explanation

The wavelength(

$\lambda$ ) of X-rays related to the energy of the electron, E, by the relation:

$\dfrac {hc}{\lambda}=E$ .
So it is inversely proportional, the minimum wavelength of emitted radiation decreases.
And the intensity of the X-rays increases linearly with decreasing frequency, from zero at the energy of the incident electrons, the voltage on the X-ray tube.

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 4 - 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 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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