cannot be sent directly over the air for large distance

can be sent directly over the air for large distance

MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 8

The photoelectric material having work-function

$\phi_{0}$ is illuminated with light of wavelength

$\lambda \left (\lambda < \dfrac {hc}{\phi_{0}}\right )$ . The fastest photoelectron has a de. Broglie wavelength

$\lambda_{d}$ . A change in wavelength of the incident light by

$\triangle \lambda$ results in change

$\triangle \lambda_{d}$ in

$\lambda_{d}$ . then the ratio

$\triangle \lambda_{d}/ \triangle \lambda$ is proportional to

0%
$\lambda_{d}/ \lambda$
0%
$\lambda_{d}^{2}/ \lambda$
0%
$\lambda_{d}^{3}/ \lambda$
0%
$\lambda_{d}^{3}/ \lambda^{2}$

Explanation

solution:

$\dfrac{hc}{\lambda } - \Phi o = \dfrac{1}{2} m v^2$

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

So,

$\dfrac{hc}{\lambda } - \Phi o = \dfrac{h^2}{2m\lambda ^2 _d}$

$\dfrac{hc \Delta \lambda }{\lambda ^2} = \dfrac{h^2 }{2 m \lambda ^3 _d} (- \Delta \lambda _d)$

$\dfrac{\Delta \lambda _d} {\Delta \lambda} \propto \dfrac{\lambda _d ^ 3 }{\lambda ^2}$

hence the correct option : D

A radiation of energy

$E$ falls normally on a perfectly reflecting surface. The momentun transferred to the surface is

0%
$E/c$
0%
$2E/c$
0%
$Ec$
0%
$E/{c}^{2}$

Explanation

Let initial momentum is ${{p}_{i}}$ and final reflected momentum is ${{p}_{f}}$ such that

Energy, $E=\dfrac{hc}{\lambda }.....(1)$

${{p}_{i}}=\dfrac{h}{\lambda }=\dfrac{E}{c}\,\,(from\,\,1)$

${{p}_{f}}=\dfrac{-h}{\lambda }=\dfrac{-E}{c}$

So, net change in momentum is

$\Delta p={{p}_{f}}-{{p}_{i}}$

$\Delta p=\dfrac{-E}{c}-\dfrac{E}{c}$

$\Delta p=-\dfrac{2E}{c}$

$\dfrac{2E}{c}$

The uncertainty in position of a particle is same as it's de Broglie wavelength, uncertainty in its momentum is _______

0%
$\cfrac { h }{ \lambda }$
0%
$\cfrac { \lambda }{ h }$
0%
$\cfrac { 2h }{ 3\lambda }$
0%
$\cfrac { 3\lambda }{ 2h }$

Explanation

$\Delta x=\lambda$

$\left( \cfrac { a }{ q } \right)$

According to the Uncertainty principle,

$\Delta x\cdot \Delta p\ge \cfrac { h }{ 4\pi }$

$\therefore \Delta p\ge \cfrac { 1 }{ 4\pi } \left( \cfrac { h }{ \lambda } \right) \\ \therefore \Delta p\propto \cfrac { h }{ \lambda }$

Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is

$0.73\ eV$ , The work function for sodium is

$1.82\ eV$ . Find the energy of the photons causing the photoelectric emissions.

0%
$2.54\ eV$ .
0%
$3.6\ eV$ .
0%
$4.7\ eV$ .
0%
$9\ eV$ .

Explanation

The energy of photon causing photoelectric emission

$=$ Work function of sodium metal

$+$

$KE$ of the fastest photoelectron

$=1.82+0.73=2.55eV$

Electrons are accelerated through a potential difference

$V$ and protons are accelerated through a potential difference

$4V$ . The de-Broglie wavelengths are

${ \lambda }_{ e }$ and

${ \lambda }_{ p }$ for electrons are protons respectively. The ratio of

$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } }$ is given by (given

${m}_{e}$ is mass of electron and

${m}_{p}$ is mass of proton)

0%
$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ p } }{ { m }_{ e } } }$
0%
$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } }$
0%
$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\cfrac { 1 }{ 2 } \sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } }$
0%
$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ e } }{ { m }_{ p } } }$

Explanation

$\cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\cfrac { \sqrt { 2{ m }_{ p }{ k }_{ p } } }{ \sqrt { 2{ m }_{ e }{ k }_{ e } } } =\cfrac { \sqrt { 2\times { m }_{ p }\times e\times 4v } }{ \sqrt { 2\times { m }_{ e }\times e\times v } } \Rightarrow \cfrac { { \lambda }_{ e } }{ { \lambda }_{ p } } =2\sqrt { \cfrac { { m }_{ p } }{ { m }_{ e } } }$

After absorbing a slowly moving neutron of mass

${m}_{N}$ (momentum

$\sim 0$ ) a nucleus of mass

$M$ breaks into two nuclei of masses

${m}_{1}$ and

$5{ m }_{ 1 }\left( 6{ m }_{ 1 }=M+{ m }_{ N } \right)$ respectively. If the de-Broglie wavelength of the nucleus with mass

${m}_{1}$ is

$\lambda$ , the de-Broglie wavelength of the other nucleus will be:

0%
$\lambda$
0%
$25\lambda$
0%
$5\lambda$
0%
$\lambda /5$

Explanation

Momentum should be conserved.

As initial momentum was

$zero$ so final should also be zero.

$p_1 +p_2=0$ or

$p_1=-p_2$ i.e.

$both$ have

$same$ values..............(1)

where

$p_1$ is momentum of

$m_1$ and

$p_2$ is that of

$m_2$

The deBroglie wavelength is given by

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

As both mass have same

$momentum,$ ,

$p_1=-p_2$

so same values of

$wavelength.$ So answer is

$\lambda$

Photon of frequency

$v$ has a momentum associated with it. If

$c$ is the velocity of light, the momentum is :

0%
$v/c$
0%
$hvc$
0%
$hv/{ c }^{ 2 }$
0%
$hv/c$

Explanation

We know,

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

$\lambda =$ wavelength

$P=\dfrac { hv }{ C }$

Thus, the momentum with

$C$ being the velocity of light

$=\dfrac { hv }{ C }$ .

If the kinetic energy of a free electron doubles, its de Broglie wavelength changes by the factor

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

Explanation

$\lambda =\frac { h }{ mv }$ and

$k=\dfrac { 1 }{ 2 } { mv }^{ 2 }=\dfrac { { \left( mv \right) }^{ 2 } }{ 2m }$

$\Rightarrow mv=\sqrt { 2mk }$

So,

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

$\Rightarrow \lambda \sim \dfrac { 1 }{ \sqrt { k } }$

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

$\left( { k }_{ 2 }=2{ k }_{ 1 } \right)$

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

What is the energy associated with photons of wavelength

$9000\ A$ ?

0%
$4.97\ \times 10^{-19}\ J$
0%
$4.97\ \times 10^{-18}\ J$
0%
$4.97\ \times 10^{-19}\ erg$
0%
none of these

The ratio of the de-Broglie wavelength of an electron to a photon is $\frac{3}{2}$ . The speed of the electron is equal to $\frac{2}{3}rd$ of a speed of light. Then the ratio of the energy of the electron to a photon is

0%
$\frac{1}{3}$
0%
$\frac{2}{9}$
0%
$\frac{3}{4}$
0%
$\frac{4}{9}$

If the stationery proton and

$\alpha-$ particle are accelerated through same potential difference, the ratio of de-Broglie wavelength will be:

0%
$2$
0%
$1$
0%
$2\surd{2}$
0%
None of these

Explanation

The gain in

$K.E$ . of a charge particle after moving through a potential difference of

$V$ is given as

$eV$ , that is also equal to

$\dfrac{1}{2}\ mv^{2}$ where

$v$ is the velocity of the charge particle.

$\dfrac{1}{2}\ mv^{2}=qV \quad \Rightarrow \quad v=\sqrt{\dfrac{2qV}{m}}$

$\therefore \quad mv=\sqrt{2mqV}$

$\Rightarrow \quad$ de-Broglie wavelength

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

$\therefore \quad \dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha} q_{\alpha} V_{\alpha}}{m_{p} q_{p} V_{p}}}$
Putting

$V_{\alpha}=V_{p},\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{(4)(2)}{(1)(1)}}=2\sqrt{2}$
Hence (

$C$ ) is correct.

The de-Broglie wavelength of an electron in first orbit of Bohr hydrogen is equal to

0%
Radius of the orbit
0%
Perimeter of the orbit
0%
Diameter of the orbit
0%
Half of the perimeter of the orbit

Explanation

The de-Broglie wavelength

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

$1$ )
Where the linear momentum of an orbiting electron is given as

$mvr=\dfrac{nh}{2\pi}$ for first orbit

$n=1$

$\Rightarrow \quad mv=\dfrac{h}{2\pi r}$ . . . . . (

$2$ )
Using (

$1$ ) and (

$2$ ) we obtain

$\lambda =\dfrac { h }{ \left( \dfrac { h }{ 2\pi r } \right) } =2\pi r$ where,

$2\pi r=$ perimeter of the orbit
Hence (

$B$ ) is correct.

Statement 1: A metallic surface is irradiated by a monochromatic light of frequency

$v> {v}_{0}$ (the threshold frequency). The maximum kinetic energy and the stopping potential are

${K}_{max}$ and

${V}_{0}$ respectively. If the frequency incident on the surface doubled, both the

${K}_{max}$ and

${V}_{0}$ are also doubled.
Statement 2- The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light..

0%
Statement- 1 is true; Statement- 2 is true; Statement- 2 is the correct explanation of Statement- 1
0%
Statement- 1 is true; Statement- 2 is true; Statement- 2 is not the correct explanation of Statement- 1
0%
Statement- 1 is false; Statement- 2 is true
0%
Statement- 1 is true; Statement- 2 is false

Explanation

${ K }_{ max }=e{ V }_{ 0 }=h\left( V-{ V }_{ 0 } \right)$

${ V }^{ 1 }=2V$

$\therefore$

${ K }_{ max }^{ 1 }=e{ V }_{ 0 }^{ 1 }=h\left( 2V-{ V }_{ 0 } \right)$

$=2{ K }_{ max }+h{ V }_{ 0 }$

$\therefore$

${ K }_{ max }^{ 1 }>2{ K }_{ max }$ and

${ V }_{ 0 }^{ 1 }>2{ V }_{ 0 }$

$\therefore$ Statement

$-1$ is false statement

$-2$ is true.

In an experiment on photoelectric effects a student plots stopping potential

${V}_{0}$ against reciprocal of the wavelength

$]lambda$ of the incident light for two different metals

$A$ and

$B$ . These are shown in the figure. Looking at the graphs, you can most appropriately say that

0%
work function of metal $B$ is greater than that of metal $A$
0%
for light certain wavelength falling on both metals. Maximum kinetic energy of electrons emitted from $A$ will be greater than other emitted from $B$
0%
work function of metal $A$ is greater than that of metal $B$
0%
Student data is not correct

Explanation

$\dfrac { hc }{ \lambda } -\phi =e{ V }_{ 0 }$

${ V }_{ 0 }=\dfrac { hc }{ e\lambda } -\dfrac { \phi }{ e }$

for metal

$A,\quad \dfrac { \phi A }{ hc } =\dfrac { 1 }{ \lambda }$

for metal

$B,\quad \dfrac { \phi B }{ hc } =\dfrac { 1 }{ \lambda }$

As the value of

$\dfrac { 1 }{ \lambda }$ (increasing & decreasing) is not specified hence we cannot say that which metal has comparatively greater or lesser work function

$\left( \phi \right)$ .

Calculate number of photon coming out per from bulb of watts. if it is 50% efficient and wavelength coming out is 600nm.

0%
$30\times { 10 }^{ 20 }$
0%
$1.5\times { 10 }^{ 20 }$
0%
$15\times { 10 }^{ 20 }$
0%
$8.5\times { 10 }^{ 25 }$

An isolated hydrogen atom emits a photon of energy

$9\ eV$ . Find momentum of the photons.

0%
$4.8 \times 10^{-23}kg-m/s$
0%
$4.8 \times 10^{-27}kg-m/s$
0%
$4.8 \times 10^{-30}kg-m/s$
0%
$7.8 \times 10^{-27}kg-m/s$

Explanation

We Know,
According to De-Broglie equation:

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

$p= \dfrac{h\times c}{\lambda \times c}= \dfrac{\textrm{energy}}{c}$

$p= \dfrac{9\times 1.6\times 10^{-19}}{3\times 10^8}$

$p=\dfrac{14.4 \times 10^{-27}}{3}$

$p=4.8\times 10^{-27}$

Hence option $\textbf B$ is correct.

The mass of photon of wavelength

$\lambda$ is given by

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

The audio signal:

0%
can be sent directly over the air for large distance
0%
cannot be sent directly over the air for large distance
0%
posses very high frequency
0%
none of above

Explanation

An Audio signal may vary from

$20Hz$ to

$20KHz$ ,

But in general, it is around

$200Hz$ only,.

This Frequency when transmitted directly to air for long-distance is attenuated

and a signal is lost in the air, That's why such frequency is first modulated to higher frequencies of order(MHz) and then transmitted to air.

Hence audio signals can't be transmitted directly to air for long distances.

A photon of energy E

$_1$ incident on a surface liberates electrons whose energy is equal to the work function W of the metal. When a photon of energy E

$_2$ is incident on the same surface, energy of the emitted electrons is 3 W. The ratio E

$_1$ :E

$_2$ is

0%
1:2
0%
2:1
0%
1:3
0%
3:1

Explanation

By photoelectric law,

$KE={ E }_{ 1 }-W\\ \Rightarrow \quad W={ E }_{ 1 }-W\\ { E }_{ 1 }=2W\\ And\quad 3W={ E }_{ 2 }-W\\ { E }_{ 2 }=4W\\ \therefore \quad { E }_{ 1 }={ E }_{ 2 }=2W:4W=1:2(a)$

$E_{1},E_{2}$ and

$E_{3}$ represent respectively the kinetic energies of an electron, an

$\alpha-$ particle and a proton each having same de-Broglies wavelength then

0%
$E_{1} > E_{2} > E_{3}$
0%
$E_{2} > E_{3} > E_{1}$
0%
$E_{1} > E_{3} > E_{2}$
0%
$E_{1} = E_{2} = E_{3}$

Let

$p$ and

$E$ denote the linear momentum and the energy of a photon. For another photon of smaller wavelength (in same medium)

0%
Both $p$ and $E$ increase
0%
$p$ increases and $E$ decreases
0%
$p$ decreases and $E$ increases
0%
Both $p$ and $E$ decrease

Explanation

De-Broglie relation:

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

and from Planks relation:

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

Therefore, as

$\lambda$ increases, both p and E increase.

Find frequency of a photon of energy

$3.1\ eV$ ?

0%
$5 \times 10^{14}\ Hz$
0%
$8 \times 10^{12}\ Hz$
0%
$8 \times 10^{14}\ Hz$
0%
$8 \times 10^{16}\ Hz$

Ultraviolet light of wavelength

$300nm$ and intensity

$1.0W{m}^{-2}$ falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of

$1.0{cm}^{2}$ of the surface is nearly

0%
$9.61\times {10}^{14}$ ${s}^{-1}$
0%
$4.12\times {10}^{13}$ ${s}^{-1}$
0%
$1.51\times {10}^{12}$ ${s}^{-1}$
0%
$2.13\times {10}^{11}$ ${s}^{-1}$

Explanation

Wavelength

$\lambda =300nm=300\times { 10 }^{ -9 }m,$

Intensity,

$I=\quad 1W{ m }^{ -2 }\\ I=\dfrac { nhc }{ \lambda } \\ \rightarrow n=\dfrac { I\lambda }{ hc } =\dfrac { 1\times 300\times { 10 }^{ -9 } }{ 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } } \\ =1.5\times { 10 }^{ 18 }\\ 1%\quad of\quad n={ { n }_{ 1 }=1.5\times { 10 }^{ 16 } }$

Now

${ { n }_{ 1 } }$ photons are incident on

$4\pi { r }^{ 2 }=1㎡$ area

$\therefore { n }^{ " }\quad photons\quad are\quad incident\quad on\quad 1{cm}^{2}={ 10 }^{ -4 }{m}^{2}\quad area\quad are\\ \dfrac { 1.5\times { 10 }^{ 16 } }{ { n }^{ " } } =\dfrac { 1 }{ { 10 }^{ -4 } } \\ \rightarrow { n }^{ " }=1.5\times { 10 }^{ 12 }/s$

The photoelectric threshold wavelength of tungsten is

$230\ nm$ . The energy of electrons ejected from its surface by ultraviolet light of wavelength

$180\ nm$ is

0%
$0.15\ eV$
0%
$1.5\ eV$
0%
$15\ eV$
0%
$15\ keV$

Explanation

Phototelectric law,

$KE=\frac { hc }{ \lambda } -W\\ =\frac { hc }{ \lambda } -\frac { hc }{ { \lambda }_{ \circ } } \\ where,\quad W=\frac { hc }{ { \lambda }_{ \circ } } \\ and\quad { \lambda }_{ \circ }\quad is\quad threshold\quad wavelength\\ \therefore \quad KE=hc\left( \frac { 1 }{ \lambda } -\dfrac { 1 }{ { \lambda }_{ \circ } } \right) \\ =\dfrac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 }\times { 10 }^{ 9 }\left( \dfrac { 1 }{ 180 } -\dfrac { 1 }{ { 230 } } \right) }{ 1.6\times { 10 }^{ -19 } } \\ =1.5eV$

The number of photons of light of wavelengths

$7000\mathring{A}$ equivalent to

$1J$ are :

0%
$3.52\times {10}^{-18}$
0%
$3.52\times {10}^{18}$
0%
$50,000$
0%
$10,0000$

Explanation

Energy,

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

Energy of a

$single$ photon of wavelength 7000 Angstrom is

$\dfrac{hc}{7000\times 10^{-10}}$

put

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

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

we get energy of a single photon as

$E= 2.829\times 10^{-19} Joule$

Number of photon in

$1\ Joule =\dfrac{1 J}{E} =3.52\times 10^{18}$

Option B is correct

$K_{1} and K_{2}$ are the maximum kinetic energies of the photo electrons emitted when light of wave lengths

$\lambda _{1} and \lambda _{2}$ respectively are incident on a metallic surface. If

$\lambda _{1}=3\lambda _{2}$ then

0%
(a) $K _{1}>\frac{K _{2}}{3}$
0%
(b) $K _{1}<\frac{K _{2}}{3}$
0%
(c) $K _{1}>3K _{2}$
0%
(d) $K _{2}=3K _{1}$

Explanation

For photoelectrons having maximum kinetic energy

$K=\dfrac { hc }{ \lambda } \\ \therefore { K }_{ 1 }=\dfrac { hc }{ { \lambda }_{ 1 } } \quad and\quad { K }_{ 2 }=\dfrac { hc }{ { \lambda }_{ 2 } } \\ Now,\quad { \lambda }_{ 1 }={ 3\lambda }_{ 2 }\\ \Rightarrow { K }_{ 1 }=\dfrac { hc }{ { 3\lambda }_{ 2 } } \\ { K }_{ 1 }=\dfrac { { K }_{ 2 } }{ { 3 } } \\ \Rightarrow { K }_{ 2 }=3{ K }_{ 1 }\quad (d)$

Let

$n_r$ and

$n_b$ be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time, then:

0%
$n_r = n_b$
0%
$n_r < n_b$
0%
$n_r > n_b$
0%
data insufficient

Explanation

$n_r \dfrac {hc}{\lambda _r}=n_b \dfrac {hc}{\lambda _b}$

$n_r=\dfrac {\lambda _r}{\lambda _b}n_b$

$\lambda _r > \lambda _b$

$\therefore \ \boxed {n_r > n_b}$

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

$\lambda_p$ and

$\lambda_e$ .

0%
$\lambda_e = \lambda_p$
0%
$\lambda_e < \lambda_p$
0%
$\lambda_e > \lambda_p$
0%
none of these

Explanation

Let

${ m }_{ e }$ and

${ m }_{ p }$ be the mass of electron and proton respectively. Let the applied potential different be

$V$ .

Thus, the de-Broglie wavelength of the

${ e }^{ - }$ ,

${ \lambda }_{ e }=\dfrac { h }{ \sqrt { 2{ m }_{ e }eV } } \quad \longrightarrow \left( 1 \right)$

and de - Broglie wavelength of proton

${ \lambda }_{ p }=\dfrac { h }{ \sqrt { 2{ m }_{ p }eV } } \quad \longrightarrow \left( 2 \right)$

Dividing,

$\dfrac { { \lambda }_{ p } }{ { \lambda }_{ e } } =\dfrac { \sqrt { { m }_{ e } } }{ \sqrt { { m }_{ p } } }$

${ m }_{ e }<{ m }_{ p }$

$\therefore$

$\dfrac { { \lambda }_{ p } }{ { \lambda }_{ e } } <1$

$\Rightarrow { \lambda }_{ p }{ \lambda }_{ e }$

The threshold wavelength of tungsten is

$2300\overset {\circ}{A}$ . If ultra violet light of wavelength

$1600\overset {\circ}{A}$ is incident on it, then the maximum kinetic energy of photoelectrons would be

0%
$1.5\ eV$
0%
$2.5\ eV$
0%
$3.0\ eV$
0%
$5.0\ eV$

The wavelength of a photon is

$4000$

$\mathring A$ . Calculate its energy:

0%
$49.5\times 10^{-19}$ $J$
0%
$495\times 10^{-19}$ $J$
0%
$4.95\times 10^{-19}$ $kJ$
0%
$4.95\times 10^{-19}$ $J$

Explanation

$\lambda=4000 \mathring A=4000\times 10^{-10}$

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

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

$E=\cfrac{6.6\times 10^{-34}\times 3\times 10^8}{4000\times 10^{-10}}$

$E=4.95 \times 10^{-19}$

$J$

de-Broglie wavelength associated with an electron revolving in the

$n^{th}$ state of hydrogen atom is directly proportional to

0%
$n$
0%
$\cfrac{1}{n}$
0%
$n^2$
0%
$\cfrac{1}{n^2}$

Explanation

$V_n = 2.165 \times 10^6(\cfrac{z}{n})$

$V_n \propto \cfrac{1}{n}$

$mV_n \propto \cfrac{1}{n}$

$\therefore \lambda = \cfrac{h}{mV_n} \propto n$

The maximum wavelength of light for photoelectric effect from a metal is

$200\ nm$ . The maximum kinetic energy of electron which is emitted by the radiation of wavelength

$100\ nm$ will be:

0%
$12.4\ eV$
0%
$6.2\ eV$
0%
$100\ eV$
0%
$200\ eV$

Explanation

Given that,

The maximum wave length $\lambda_{0} =200\,nm$

Radiation of wave length $\lambda =100\,nm$

We know that, the maximum K.E is

$K.E=12400\left[ \dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{0}}} \right]$

$K.E=12400\left[ \dfrac{1}{1000}-\dfrac{1}{2000} \right]$

$K.E=12400\times \dfrac{1}{2000}$

$K.E=6.2\,eV$

Hence, the kinetic energy is $6.2\ eV$

The de-Brogile wavelength corresponding to the root mean square velocity of the hydrogen molecule at

$20^{o}C$ .

0%
$2.04\ \mathring {A}$
0%
$1.04\ \mathring {A}$
0%
$3.04\ \mathring {A}$
0%
$4.04\ \mathring {A}$

Explanation

Given that,

$T=20+273=293\,K$

Molecular mass of ${{H}_{2}}=2\times 1.00794=2.01588\times 1.66\times {{10}^{-27}}\,kg$

We know that,

$\dfrac{1}{2}mv_{rms}^{2}=\dfrac{3}{2}kT$

${{v}_{rms}}=\sqrt{\dfrac{3kt}{m}}$

Now, the de Broglie wave length

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

$\lambda =\dfrac{h}{\sqrt{3mkT}}$

$\lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{3\times 2.01588\times 1.66\times {{10}^{-27}}\times 1.38\times {{10}^{-23}}\times 293}}$

$\lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{4059.20\times {{10}^{-50}}}}$

$\lambda =\dfrac{6.63\times {{10}^{-34}}}{63.712\times {{10}^{-25}}}$

$\lambda =0.104\times {{10}^{-9}}$

$\lambda =1.04\overset{\circ }{\mathop{A}}\,$

Hence, the de Broglie wave length is $1.04\overset{\circ }{\mathop{A}}\,$

if the frequency of the incident light falling on a photosensitive material is doubled, then

$K.E$ . Of the emitted photoelectrons:

0%
remains constant
0%
becomes two times its initial value
0%
becomes more then two times its initial value
0%
becomes less then two times its initial value

Explanation

Kinetic energy of photoelectron is ${{E}_{1}}$ and ${{E}_{2}}$ for incident light of frequency $\upsilon$ and $2\upsilon$ respectively

Now, $h\upsilon ={{E}_{1}}+{{\phi }_{0}}.....(I)$

$2h\upsilon ={{E}_{2}}+{{\phi }_{0}}.....(II)$

Now, put the value of $h\upsilon$ in equation (II)

$2\left( {{E}_{1}}+{{\phi }_{o}} \right)={{E}_{2}}+{{\phi }_{0}}$

${{E}_{2}}=2{{E}_{1}}+{{\phi }_{0}}$

So, the kinetic energy of the emitted photoelectrons becomes more then two times its initial value

Electron has energy of 100 eV what will be its wavelength

0%
$1.2 \mathring { A }$
0%
$10 \mathring { A }$
0%
$100 \mathring { A }$
0%
$1 \mathring { A }$

Explanation

Energy

$E=100ev=100\times 1.6 \times 10^{-19} Joule$

Wavelength,

$\lambda = \dfrac{h}{\sqrt[2]{2mE}}=\dfrac{6.62\times 10^{-34}Js}{\sqrt[2]{2\times 9.1\times 10^{-31} Kg \times 1.6\times 10^{-17}J}}=1.226\times 10^{-10} m=1.2Angstrom$

Option A is correct.

In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper

$(\phi = 4.5 eV)$ . the emitter is illuminated by a source of monochromatic light of wavelength 200 nm.

0%
The minimum kinetic energy of the photo electrons reaching the collector is 0
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The maximum kinetic energy of the photo electrons reaching collector is 3.7 eV.
0%
If the polarity of the battery is reversed then answer to part A will be 0
0%
If the polarity of the battery is reversed then answer to part B will be 1.7 eV

Explanation

$K = \displaystyle {{hc} \over \alpha } - \phi$

$= \displaystyle {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {200 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}} - 4.5$

$= 1.7eV$

Min. energy with which electrons emitted=1.7eV

Max. $K\varepsilon \Rightarrow {K_{\min }} + e{V_0}$

$= 1.7 + 2$

$= 3.7eV$

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

${ \lambda }_{ 1 }$ be de-Broglie wavelength of the proton and

${ \lambda }_{ 2 }$ be the wavelength of the photon.The ration

$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } }$ is proportional to:

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

Explanation

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

photon energy

$E=hv$

So,

${ \lambda }_{ 1 }=\frac { h }{ p } =\frac { h }{ \sqrt { 2mE } } \quad \longrightarrow \left( i \right) \quad \Rightarrow E=h{ v }_{ 2 }=\dfrac { { h }_{ c } }{ { \lambda }_{ 2 } }$

${ \lambda }_{ 2 }=\dfrac { { h }_{ c } }{ E } \quad \longrightarrow \left( ii \right)$ from eq

$(i)$ &

$(ii)$

$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\dfrac { h\sqrt { 2mE } }{ \left( \dfrac { { h }_{ c } }{ E } \right) } \sim \dfrac { E }{ \sqrt { E } } ={ E }^{ 1/2 }$

Electrons used in an electron microscope are accelerated by a voltage of

$25$ kV. If the voltage is increased to

$100$ kV then the de-Broglie wavelength associated with the electrons would?

0%
Increase by $2$ times
0%
Decrease by $2$ times
0%
Decrease by $4$ times
0%
Increase by $4$ times

Explanation

De Broglie wavelength in terms of accelerated potential difference is

$\lambda =\dfrac{12.27}{\sqrt{V}}$

Initial and final wavelength are ${{\lambda }_{1}}$ and ${{\lambda }_{2}}$ having potential as ${{V}_{1}}$ and ${{V}_{2}}$

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

$\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\dfrac{100}{25}}$

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

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

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

So, new wavelength is decrease by 2.

When a photon of light collides with a metal surface, number of electrons, (if any) coming out is

0%
only one
0%
only two
0%
infinite
0%
depends upon factors

$10^{-3}W$ of

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

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

0%
$0.4\%$
0%
$.04\%$
0%
$20\%$
0%
$10\%$

A photo sensitive surface is receiving light of wavelength

$6000 \mathring{A}$ at the rate of

$10^{-7} J/s$ . The number of photons striking the surface per second is approximately equal to:

0%
$1.5 \times 10^{11}$
0%
$3 \times 10^{11}$
0%
$4 \times 10^{11}$
0%
$6 \times 10^{11}$

By applying a force of one Newton, one can hold a body of mass-

0%
102 grams
0%
102 kg
0%
102 mg
0%
None of these

If light of wavelength

$6600A^{\circ}$ is incident on a metal of work function

$2eV$ , then the maximum kinetic energy of emitted photo electrons will be-

0%
2 eV
0%
1 eV
0%
0.5 eV
0%
Emission pf photo electrons will not take place

Which of the following photon of light have higher energy than photon of green light?

0%
Red
0%
Violet
0%
Orange
0%
None of these

Explanation

Energy of the photon depends on the frequency of the incident light it is directly proportional to the frequency and inversely proportional to the wavelength. As in visible light spectrum frequency increases in the order

$red<yellow<orange<green<blue<indigo<violet$ . Violet light has greater frequency than green light hence it will have greater energy.

What is de-Broglie wavelength of electron having energy

$10\ keV$ ?

0%
$0.12$ $A^{\circ}$
0%
$1.2$ $A^{\circ}$
0%
$12.2$ $A^{\circ}$
0%
none of these.

The de-Broglie wavelength of an electron in the first Bohr orbit is

0%
Equal to one fourth the circumference of the first orbit
0%
Equal to twice the circumference of the first orbit
0%
Equal to half the circumference of the first orbit
0%
Equal to the circumference of the first orbit

Explanation

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

$\Rightarrow pr_n=\dfrac{nh}{2\pi}$

$\Rightarrow \dfrac{h}{\lambda}\times r_n=\dfrac{nh}{2\pi}$

$\Rightarrow \lambda =\dfrac{2\pi r_n}{n}$ , for first orbit

$n=1$ so

$\lambda =2\pi r_1$
= circumference of first orbit

In a photoelectric experiment, with light of wavelength

$\lambda$ , the fastest electron has speed

$v$ . If the wavelength is changed to

$\dfrac{3\lambda}{4}$ , the speed of the fasted emitted electron will become

0%
$v\sqrt{\dfrac{3}{4}}$
0%
$v\sqrt{\dfrac{4}{3}}$
0%
less than $v\sqrt{\dfrac{3}{4}}$
0%
greater than $v\sqrt{\dfrac{4}{3}}$

In a photo emissive all with exciting wavelength

$\lambda$ , the fastest electron has a speed

$v$ . If the wavelength is changed to

$\dfrac {3}{4}\lambda$ , the speed of the fastest emitted electron will be

0%
$V\sqrt {\dfrac {3}{4}}$
0%
$V\sqrt {\dfrac {4}{3}}$
0%
Less than $V\sqrt {\dfrac {4}{3}}$
0%
More than $V\sqrt {\dfrac {4}{3}}$

Photoelectric effect takes place in element A. Its work function is

$2.5\ eV$ and threshold wavelength is

$\lambda$ . An other element

$B$ is having work function of

$5\ eV$ . Then find out the wavelength that can produce photoelectric effect in

$B$

0%
$\dfrac{\lambda}{2}$
0%
$2\lambda$
0%
$\lambda$
0%
$3\lambda$

Explanation

We know the relation between work function,

$W$ and threshold wavelength,

$\lambda$ is

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

$\lambda \propto \dfrac{1}{W}$

Using above relation we notice that work function for

$B$ is

$double$ of that for

$A$ so the threshold wavelength for

$B$ will be

$half$ of that for

$A$ that is answer is

$\dfrac{\lambda }{2}$

The approximate wavelength of a photon of energy

$2.48$ eV is

0%
$500 \mathring{A}$
0%
$5000 \mathring{A}$
0%
$2000 \mathring{A}$
0%
$1000 \mathring{A}$

Explanation

We know that,

$E\left( eV \right) =\dfrac { 12375 }{ \lambda \overset { 0 }{ A } }$

$\Rightarrow \lambda =\dfrac { 12375 }{ \left( 2.48 \right) }$

$\lambda =4989.9\overset { 0 }{ A } \approx 5000\overset { 0 }{ A }$

Approximate wavelength of a photon of energy

$2.48eV=5000\overset { 0 }{ A }$ .

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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