MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 7

The minimum intensity of light to be detected by human eye is

${ 10 }^{ -10 }W/{ m }^{ 2 }$ . The number of photons of wavelength

$5.6\times { 10 }^{ -7 }m$ entering the eye, with pupil area

${ 10 }^{ -6 }{ m }^{ 2 }$ , per second for vision will be nearly

0%
$100$
0%
$200$
0%
$300$
0%
$400$

Explanation

On using

$l=\cfrac { P }{ A }$ , where

$P=$ radiation power

$\quad \Rightarrow P=l\times A=\cfrac { nhc }{ t\lambda } =lA\Rightarrow \cfrac { n }{ t } =\cfrac { lA\lambda }{ hc }$
Hence the number of photons entering per second
eye

$\left( \cfrac { n }{ t } \right) =\cfrac { { 10 }^{ -10 }\times { 10 }^{ -6 }\times 5.6\times { 10 }^{ -7 } }{ 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } } =300$

If alpha particle, proton and electron move with the same momentum, then their respective de-Broglie wavelengths

${ \lambda }_{ \alpha },{ \lambda }_{ p },{ \lambda }_{ e }$ are related as

0%
${ \lambda }_{ \alpha }={ \lambda }_{ p }={ \lambda }_{ e }$
0%
${ \lambda }_{ \alpha }<{ \lambda }_{ p }<{ \lambda }_{ e }$
0%
${ \lambda }_{ \alpha }>{ \lambda }_{ p }>{ \lambda }_{ e }$
0%
${ \lambda }_{ p }>{ \lambda }_{ e }>{ \lambda }_{ \alpha }$
0%
${ \lambda }_{ p }<{ \lambda }_{ e }<{ \lambda }_{ \alpha }$

Explanation

Wavelength

$\lambda =\dfrac { h }{ p }$
Since, momentum is same for all particles.
So, wavelength will be same for all
i.e.,

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

De-Broglie wavelength of a body of mass

$1$ kg moving with velocity of

$2000 \>$ m

$/$ s is?

0%
$3.32\times 10^{-27}\overset{o}{A}$
0%
$1.5\times 10^7\overset{o}{A}$
0%
$0.55\times 10^{-22}\overset{o}{A}$
0%
None of these

Explanation

de Broglie wavelength

$\lambda \displaystyle =\frac{h}{mv}=\frac{6.6\times 10^{-34}}{1\times 2000}$

$=3.3\times 10^{-37}m=3.3\times 10^{-27}\overset{o}{A}$ .

In a photoemissive cell with exciting wave length

$\lambda$ , the fastest electron has a speed

$v$ . If the exciting wavelength is change to

$3\lambda / 4$ , then the speed of the fastest emitted electron will be

0%
$v\left (\dfrac {3}{4}\right )^{\dfrac {1}{2}}$
0%
$v\left (\dfrac {4}{3}\right )^{\dfrac {1}{2}}$
0%
Less than $v\left (\dfrac {4}{3}\right )^{\dfrac {1}{2}}$
0%
Greater than $v\left (\dfrac {4}{3}\right )^{\dfrac {1}{2}}$

Explanation

From Einstein's photoelectric equation,

$\dfrac {1}{2}mv^{2} = \dfrac {hc}{\lambda} - W .... (i)$

where

$W$ is the work function of metal.
For case 2 :

$\dfrac {1}{2}m(v')^{2} = \dfrac {hc}{3\lambda / 4} - W .... (ii)$
On dividing Eq. (ii) by Eq. (i), we get

$\left (\dfrac {v'}{v}\right )^{2} = \dfrac {\dfrac {4hc}{3\lambda} - W}{\dfrac {hc}{\lambda} - W}$

$\because W > 0$

$\implies \ v' > v\left (\dfrac {4}{3}\right )^{\dfrac {1}{2}}$ .

The de-Broglie wavelength of a neutron at

${ 27 }^{ o }C$ is

$\lambda$ . What will be its wavelength at

${ 927 }^{ o }C$ ?

0%
$\lambda /2$
0%
$\lambda /3$
0%
$\lambda /4$
0%
$\lambda /9$

Explanation

$\quad { \lambda }_{ neutron }\propto \cfrac { 1 }{ \sqrt { T } } \Rightarrow \cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\sqrt { \cfrac { { T }_{ 2 } }{ { T }_{ 1 } } }$

$\Rightarrow \cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\sqrt { \cfrac { (273+927) }{ (273+27) } } =2\Rightarrow { \lambda }_{ 2 }=\cfrac { \lambda }{ 2 }$

Threshold wavelength of a metal is

$4000\overset{o}{A}$ . If light of wavelength

$3000\overset{o}{A}$ irradiates the surface, the maximum kinetic energy of photoelectron is?

0%
$1.7$ eV
0%
$1.6$ eV
0%
$1.5$ eV
0%
$1.0$ eV

Explanation

If the maximu kinetic energy of photoelectron emitted from the surface of a metal is

$E_k$ and

$W$ is work function of the metal, the from Einstein photoelectric equation we have

$E_k=hv-W$
where, hv is the energy of the photon absorbed the electron in the metal.
Also

$v=\displaystyle\frac{c}{\lambda}$

$=\displaystyle\frac{velocity}{wavelength}$

$=E_k=\displaystyle\frac{hc}{\lambda}-\frac{hc}{\lambda_0}=hc\left(\displaystyle\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$
Given,

$\lambda =3\times 10^{-7}$ m,

$\lambda_0=4\times 10^{-7}$ m.

$\therefore E_k=.6\times 10^{-34}\times 3\times 10^8\times \left(\displaystyle\frac{1}{3\times 10^{-7}}-\frac{1}{4\times 10^{-7}}\right)$

$=\displaystyle\frac{19.8\times 10{-19}}{12\times 1.6\times 10^{-19}}$ eV

$=1.03\approx 1$ eV.

The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is (c =velocity of light, h= Planck's constant)

0%
$h$
0%
$c$
0%
$\dfrac {1}{c}$
0%
None of these

Explanation

We have

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

and

$\lambda_p=\dfrac {h}{mc}$

According to the question,

$\lambda_e=\lambda_p$

$\therefore v=c$

$\dfrac {E_p}{P_e} =\dfrac {mc^2}{mv}= \dfrac {c^2}{c}= c$

The speed of an electron having a wavelength of

${ 10 }^{ -10 }m$ is

0%
$4.24\times { 10 }^{ 6 }m/s\quad$
0%
$5.25\times { 10 }^{ 6 }m/s$
0%
$6.26\times { 10 }^{ 6 }m/s\quad$
0%
$7.25\times { 10 }^{ 6 }m/s\quad$

Explanation

We have

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

$\Rightarrow v=\cfrac { h }{ m\lambda } =\cfrac { 6.63\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times { 10 }^{ -10 } } =7.25\times { 10 }^{ 6 }m/s\quad$

If alpha particle and deutron move with velocity

$v$ and

$2v$ respectively, the ratio of their de-Broglie wavelength will be _____

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

Explanation

Key Concept According to de-Broglie wavelength, the wave associated with moving particle is given by

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

$m$ and

$v$ are the mass and velocity of the particle and

$h$ is Planck's constant.
de-Broglie wavelength of a particle, is given as

$\lambda = \dfrac {h}{P}\Rightarrow \lambda \propto \dfrac {1}{P}$

$\Rightarrow \dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {P_{2}}{P_{1}}$ or

$\dfrac {\lambda_{\alpha}}{\lambda_{D}} = \dfrac {P_{D}}{P_{1}}$

$\Rightarrow \dfrac {\lambda_{\alpha}}{\lambda_{D}} = \dfrac {2\times 2v}{4\times 2v} = \dfrac {4v}{4v}$

$\therefore \lambda_{\alpha} : \lambda_{D} = 1 : 1$ .

The energies of the incident photons are $3,4,5\,eV.$ The work functions of the metals are $0.5,1.5,2.5\,eV.$ The maximum K.E.s of the photoelectrons are in the ratio:

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

Explanation

${ E }_{ 1 }=3eV\\ { E }_{ 2 }=4eV\\ { E }_{ 3 }=5eV\\ W_{ 1 }=0.5eV\\ { W }_{ 2 }=1.5eV\\ { W }_{ 3 }=2.5eV\\$

From photoelectric law:

$KE=\frac { 1 }{ 2 } m{ v }^{ 2 }=E-W\\ \therefore { KE }_{ 1 }=3-0.5\\ =2.5eV\\ { KE }_{ 1 }=4-1.5\\ =2.5eV\\ { KE }_{ 1 }=5-2.5\\ =2.5eV\\ \therefore { KE }_{ 1 }:{ KE }_{ 2 }:{ KE }_{ 3 }=1:1:1$

If the kinetic energy of free electron is made double; the new de-Broglie wavelength will be __________ times that of initial wavelength.

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

Explanation

We know that
de-Broglie wavelength

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

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

$K$ be the kinetic energy of the electron, then

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

$v=\sqrt { { 2K }/{ m } }$
So, de-Broglie wavelength is

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

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

$\therefore { \lambda }_{ 1 }=\dfrac { 1 }{ \sqrt { { K }_{ 1 } } }$ ....(i)

${ \lambda }_{ 2 }=\dfrac { 1 }{ \sqrt { 2{ K }_{ 1 } } }$ ....(ii)

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

$\dfrac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 } } =\dfrac { 1 }{ \sqrt { 2 } }$
So, the new de-Broglie wavelength will be

$\dfrac { 1 }{ \sqrt { 2 } }$ times that of initial wavelength.

Light of wavelength

$\lambda$ falls on a metal having work function

$\cfrac { hc }{ { \lambda }_{ 0 } }$ . Photoelectric effect will take place only if

0%
$\lambda \ge { \lambda }_{ 0 }$
0%
$\lambda \ge 2{ \lambda }_{ 0 }$
0%
$\lambda \le { \lambda }_{ 0 }$
0%
$\lambda =4{ \lambda }_{ 0 }$

A body of mass

$100 g$ moves at the speed of

$36 { km }/{ h }$ . The de-Broglie wavelength related to it is of the order _________

$m$ .

$\left( h=6.626\times { 10 }^{ -34 }Js \right)$

0%
${ 10 }^{ -24 }$
0%
${ 10 }^{ -14 }$
0%
${ 10 }^{ -34 }$
0%
${ 10 }^{ -44 }$

Explanation

Given,

$m=100 g=0.1 kg$

$v=36{ km }/{ h }=36\times \dfrac { 5 }{ 18 } { m }/{ s }=2\times 5=10{ m }/{ s }$

$h=6.626\times { 10 }^{ -34 }Js$
We know that

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

$\left( \because p=\dfrac { h }{ \lambda } \quad and\quad p=mv \right)$

$=\dfrac { 6.626\times { 10 }^{ -34 } }{ 0.1\times 10 }$

$\lambda =6.626\times { 10 }^{ -34 }m$
So, the de-Broglie wavelength related to it is of the order

${ 10 }^{ -34 }m$

If the linear momentum of a particle is

$2.2 \times 10^4\, kg\, ms^{-1}$ , then what will be its de-Broglie wavelength ? (Take hr.

$6.6 \times 10^{-34}\, Js$ )

0%
$3 \times 10^{-29}\, m$
0%
$3 \times 10^{-29}\, nm$
0%
$6 \times 10^{-29}\, m$
0%
$6 \times 10^{-29}\, nm$

Explanation

Given, the linear momentum of aparticular (p)

$= 2.2 \times 10^4\, kg \, ms^{-1}$

$h= 6.6 \times 10^{-34} \, Js$
The de-Broglie wavelength of particle

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

$\lambda =\dfrac {6.6 \times 10^{-34}}{2.2 \times 10^4}$
or

$\lambda = 3\times 10^{-38}\, m$
or

$\lambda = 3\times 10^{-29}\, nm$

The wavelength

$\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass m, speed of light c and planck constant.

0%
$\displaystyle\frac{\lambda mc}{h}$
0%
$\displaystyle\frac{hmc}{\lambda}$
0%
$\displaystyle\frac{2hmc}{\lambda}$
0%
$\displaystyle\frac{2\lambda mc}{h}$

Explanation

The de-Broglie wavelength

$h=\displaystyle\frac{h}{mv}\Rightarrow v=\frac{h}{m\lambda}$ .........(i)
Energy of photon

$E_p=\displaystyle\frac{hc}{\lambda}$ (since

$\lambda$ is same) ...........(ii)
Energy of Photon
Kinetic energy of photon

$=\displaystyle\frac{E_p}{E_e}=\frac{hc/\lambda}{\displaystyle\frac{1}{2}1mu^2}=\frac{2hc}{\lambda mv^2}$
Substituting value of v from Eq. (i), we get

$\displaystyle\frac{E_p}{E_e}=\frac{2hc}{\lambda_m\left(\displaystyle\frac{h}{m\lambda}\right)^2}=\frac{2\lambda mc}{h}$ .

What is the photon flux and photon density at 2 m from the lamp?

0%
$5.9\times 10^{14} photons cm^{-2}, 2\times 10^4 photons cm^{-2}$
0%
$2\times 10^4 photons cm^{-2}, 5.9\times 10^{14} photons cm^{-2}$
0%
$5.9\times 10^{10} photons cm^{-2}, 2\times 10^3 photons cm^{-2}$
0%
$2\times 10^3 photons cm^{-2}, 5.9\times 10^{10} photons cm^{-2}$

Explanation

$P=100W,\quad \lambda =5890{ A }^{ 0 }$

Intensity,

$I=\dfrac { P }{ A } =\dfrac { 100 }{ 4\pi { r }^{ 2 } } =\dfrac { 100 }{ 4\times \pi \times 4 } =1.98W/㎡\\ also,\quad I=\dfrac { nhc }{ \lambda 4\pi { r }^{ 2 }t } \\ \therefore photonflux=\dfrac { n }{ 4\pi { r }^{ 2 }t } =\dfrac { I\times \lambda }{ hc } \\ =\dfrac { 1.98\times 5890\times { 10 }^{ -10 } }{ 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } } \\ =5.91\times { 10 }^{ 18 }\times { 10 }^{ -4 }\\ =5.91\times { 10 }^{ 14 }photon/㎠$

An electron in an electron microscope with initial velocity

$v_0i$ enters a region of a stray transverse electric field

$E_0j$ . The time taken for the change in its de Broglie wavelength from the initial value of

$\lambda$ to

$\dfrac{\lambda}{3}$ is proportional to.

0%
$E_0$
0%
$\displaystyle\frac{1}{E_0}$
0%
$\displaystyle\frac{1}{\sqrt{E_0}}$
0%
$\displaystyle\sqrt{E_0}$

Explanation

Path traced by the electron is shown in the figure such that electron reaches the point P when its de Broglie wavelength becomes

$\dfrac{\lambda}{3}$ .
Acceleration of electron is zero in x direction.
Acceleration of electron in y direction

$a' = \dfrac{eE_o}{m}$
Velocity of electron initially

$v_o = \dfrac{h}{\lambda}$
At P,

$v_p = \dfrac{h}{(\lambda/3)} = 3v_o$
So, velocity is y direction

$v' = \sqrt{(v')^2 - v_o^2} = \sqrt{9v_o^2-v_o^2} = 2\sqrt{2}v_o$
Using

$v' =a't$

$\therefore$

$2\sqrt{2}v_o = \dfrac{eE_o}{m}t$
We get

$t = \dfrac{2\sqrt{2}v_o m}{eE_o}$

$\implies \ t\propto \dfrac{1}{E_o}$

The approximate number of photons in a femtosecond

$(10^{-15}s)$ pulse of 600 nanometers wavelenghth light from a 10-kilowatt peak-power dye laser is

0%
$10^3$
0%
$10^7$
0%
$10^{11}$
0%
$10^{15}$
0%
$10^{18}$

Explanation

Number of photon,

$n = \dfrac{Total\ Energy}{Energy\ due\ to\ a\ single\ photon}$

Energy of a single photon,

$E_{single} = \dfrac{hc}{\lambda}$

$h = 6.63 \times 10^{−34} Js$

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

$λ = 600\ nanometers = 600 \times 10^{−9} \ m$

Energy due to a single photon =

$\dfrac{6.63 \times 10^{−34} \times 3 \times 10^8}{600 \times 10^{−9}} = 0.033 \times 10^{−17}$

Power,

$P = \dfrac{Total\ Energy}{Time}$

$Total\ energy = P \times t$

Given,

$P = 10 kW = 10^4\ W$

$t = 10^{−15} sec$

$Total\ energy = 10^4 \times 10^{−15} = 10^{-11} \ J$

$n = \dfrac{Total\ Energy}{Energy\ due\ to\ a\ single\ photon} = \dfrac{10^{−11}}{0.033 \times 10^{−17}} = 30.30 × 10^{6} = 3 × 10^7$

$\therefore n = 10^7 \ photons\ (approx.)$

Hence, the correct answer is OPTION B.

An electron and a photon have same wavelength of

$10^{-9}$ m. If E is the energy of the photon and p is the momentum of the electron, the magnitude of E/p in SI units is?

0%
$1.00\times 10^{-9}$
0%
$1.50\times 10^8$
0%
$3.00\times 10^8$
0%
$1.20\times 10^7$

Explanation

Given:-

Wavelength of photon & electron are equal

$\Rightarrow { \lambda }_{ p }={ \lambda }_{ e }={ 10 }^{ -9 }m$

$E$ =Energy of photon

$P$ = Momentum of electron

Solution:-

From Planck's equation

$E=\cfrac { hc }{ { \lambda }_{ p } } \Rightarrow { \lambda }_{ p }=\cfrac { hc }{ E }$

From de-broglie's equation

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

Since

${ \lambda }_{ p }={ \lambda }_{ e }\\ \Rightarrow \cfrac { hc }{ E } =\cfrac { h }{ p }$

$\Rightarrow \cfrac { E }{ p } =C=3\times { 10 }^{ 8 }$

The energy of a

$K$ -electron in tungsten is

$-20keV$ and of an

$L$ -electrons is

$-2keV$ . The wavelength of X-rays emitted when there is electron jump from

$L$ to

$K$ shell:

0%
$0.3443\mathring { A }$
0%
$0.6887\mathring { A }$
0%
$1.3982\mathring { A }$
0%
$2.78\mathring { A }$

Explanation

$E_{L} = - 2KeV =-2×10^{3}×1.6×10^{-19}J$

$E_{K} = - 20KeV =-20×10^{3}×1.6×10^{-19}J$

The energy of the x-ray photon is given as

$E = E_{L}-E_{K}= hv =\dfrac{hc}{\lambda }$

Or,

$\lambda = \dfrac{hc}{E_{L}-E_{K}} =\dfrac{6.62×10^{-34}×3×10^{8}}{18×10^{3}1.6×10^{-19}}m = 0.6887\dot{A}$

The formula of kinetic mass of photon is? Where h is Planck's constant, v is frequency of the photon and c is its speed.

0%
$\displaystyle\frac{hv}{c}$
0%
$\displaystyle\frac{hv}{c^2}$
0%
$\displaystyle\frac{hc}{v}$
0%
$\displaystyle\frac{c^2}{hv}$

Explanation

Energy of photon having frequency

$v$ is given by

$E = hv$
Energy of photon can be written as

$E = m_k c^2$
where

$m_k$ is the kinetic mass of photon

$\therefore$

$m_k c^2 = hv$

$\implies \ m_k = \dfrac{hv}{c^2}$

A particle is projected horizontally with a velocity

$10\ m/s$ . What will be the ratio of de-Broglie wavelengths of the particle, when the velocity vector makes an angle

$30^{\circ}$ and

$60^{\circ}$ with the horizontal?

0%
$\sqrt {3} : 1$
0%
$1 : \sqrt {3}$
0%
$2 : \sqrt {3}$
0%
$\sqrt {3} : 2$

Explanation

Given:-

${ V }_{ 1 }=10{ m }/{ s }$

Solution:-

Velocity when velocity vector makes

${30}^{o}$ with

horizontal,

${ V }_{ 2 }={ V }_{ 1 }\times \cos{ 30 }^{o}$

$={ V }_{ 1 }\times \cfrac { \sqrt { 3 } }{ 2 }$

${ V }_{ 2 }=\cfrac { 10\sqrt { 3 } }{ 2 } =5\sqrt { 3 } { m }/{ s }$

Velocity when velocity vector makes

${ 60 }^{o}$ with

horizontal,

${ { V }_{ 3 } }={ V }_{ 1 }\cos{ 60 }^{ o }$

${ V }_{ 3 }={ V }_{ 1 }\times \cfrac { 1 }{ 2 }$

${ V }_{ 3 }=10\times \cfrac { 1 }{ 2 }$

${ V }_{ 3 }=5{ m }/{ s }$

$\therefore$ Ratio of wavelength is:-

$\sqrt { 3 } :1$

de-Broglie wavelength of atom at

$TK$ absolute temperature will be

0%
$\cfrac { h }{ mKT }$
0%
$\cfrac { h }{ \sqrt { 3mKT } }$
0%
$\cfrac { \sqrt { 2mKT } }{ h }$
0%
$\sqrt { 2mKT }$

Explanation

de-Broglie wavelength

$\lambda = \dfrac{h}{mv}$
Kinetic energy of the atom

$K.E = \dfrac{1}{2}mv^2 = \dfrac{3}{2}KT$

$\implies \ mv = \sqrt{3mKT}$
Thus we get

$\lambda = \dfrac{h}{\sqrt{3mKT}}$
So, option B is correct.

The momentum associated with photon is given by __________.

0%
$h \upsilon$
0%
$\displaystyle \frac{h \upsilon}{c}$
0%
$h E$
0%
$h \lambda$

Explanation

By Einstein's formula energy of any particle is given by

$E=mc^2$

By Planck's theory the energy of a photon is given by

$E=h\upsilon$

Equating both, we get

$mc^2=h\upsilon$

But momentum

$=p=mc=\cfrac{h\upsilon}c$

An electron of mass m with an initial velocity

$\overrightarrow{V}=V_0\hat{i}(V_0 > 0)$ enters an electric field

$\overrightarrow{E}=-E_0\hat{i}$ (

$E_0=$ constant

$> 0$ ) at

$t=0$ . If

$\lambda_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is?

0%
$\lambda_0\left(\displaystyle 1+\frac{eE_0}{mV_0}t\right)$
0%
$\lambda_0 t$
0%
$\displaystyle\frac{\lambda_0}{\left(\displaystyle 1+\frac{eE_0}{mV_0}t\right)}$
0%
$\lambda_0$

Explanation

R.E.F image

The diagram will be :-

Initial velocity

$= V_{0}$

$\because$ Acceleration is constant :-

Final velocity

$= v = v_{0}+\dfrac{eE_{0}}{m}t$

(at time t)

Now, at

$t = 0 \Rightarrow$ de-Broglie wavelength

$= \lambda _{0}$

$\Rightarrow \lambda _{0} = \dfrac{h}{mv_{0}}$

Let at time t, de-Broglie wavelength

$= \lambda$

Then,

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

$= \dfrac{h}{m(V_{0}+\dfrac{eE_{0}t}{m})}$

$= \dfrac{h}{mV_{0}+eE_{0}t}$

$= \dfrac{1}{1/\lambda _{0}+eE_{0}t/h}$

$= \dfrac{\lambda .h}{h+eE_{0}\lambda _{0}t}$

$= \dfrac{\lambda _{0}}{1+\dfrac{eE_{o}\lambda _{0}t}{h}}$

$= \boxed{\lambda = \dfrac{\lambda _{0}}{1+(\dfrac{eE_{0}}{mV_{0}})t}}$

1123356_885644_ans_a08e842334cc4d6caed95f9e3f41389c.png

A charged particle is accelerated from rest through a certain potential difference. The de Broglie wavelength is

${ \lambda }_{ 1 }$ when it is accelerated through

${V}_{1}$ and is

${ \lambda }_{ 2 }$ when accelerated through

${V}_{2}$ . The ratio

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

0%
${ V }_{ 1 }^{ 3/2 }:{ V }_{ 2 }^{ 3/2 }$
0%
${ V }_{ 2 }^{ 1/2 }:{ V }_{ 1 }^{ 1/2 }$
0%
${ V }_{ 1 }^{ \cfrac { 1 }{ 2 } }:{ V }_{ 2 }^{ \cfrac { 1 }{ 2 } }$
0%
${ V }_{ 1 }^{ 2 }:{ V }_{ 2 }$

Explanation

de Broglie wavelength is given by

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

$p = \sqrt{2mK}$ and kinetic energy

$K =qV$

$\implies \ \lambda = \dfrac{h}{\sqrt{2mqV}}$
We get

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

$\implies\ \lambda_1:\lambda_2 = V_2^{1/2}:V_1^{1/2}$

An electron of mass m and a photon have same energy E.Find out the ratio of de-Brogile wavelength associated with them is (c- velocity of light)

0%
$c { \left[ 2mE \right] }^{ \dfrac { 1 }{ 2 } }$
0%
$\dfrac { 1 }{ c } { \left[ \dfrac { 2m }{ E } \right] }^{ \dfrac { 1 }{ 2 } }$
0%
$\dfrac { 1 }{ c } { \left[ \dfrac { E }{ 2m } \right] }^{ \dfrac { 1 }{ 2 } }$
0%
${ \left[ \dfrac { E }{ 2m } \right] }^{ \dfrac { 1 }{ 2 } }$

A material particle with a rest mass

$m_0$ is moving with a velocity of light c. Then determine the wavelength of the de Broglie wave associated with it is.

0%
$\displaystyle(h/m_0c)$
0%
zero
0%
$\displaystyle\infty$
0%
$\displaystyle(m_0c/h)$

Explanation

The de Broglie wavelength is

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

Where

$h$ = planks constant

$m$ = rest mass

$v$ = Velocity

Here

$m_{o}$ is the rest mass and

$c$ is the velocity. So the de Broglie wavelength is

$\lambda =\frac{h}{m_{o}c}$

The frequencies of X-rays,

$\gamma$ rays and Ultra violet rays are respectively incident on a metal, having work function is

$0.5eV$ . The ratio of maximum speed of emtted electrons is

0%
$p< q, q> r$
0%
$p> q, q >r$
0%
$p< q, q< r$
0%
$p> q, q< r$

Explanation

Maximum speed of emitted electrons

$\dfrac{1}{2}mv^2 = h\nu_{incident} - \phi_{metal}$
where

$\phi_{metal} = 0.5eV$ is the work function of the metal and

$\nu_{incident}$ is the frequency of incident radiation.
Increasing order of frequency : Ultra violet rays < X-rays <

$\gamma$ rays
Thus increasing order of maximum speed of emitted electrons : Ultra violet rays < X-rays <

$\gamma$ rays

$\implies \ r<p<q$
So option A is correct.

About 5% of the power of a 100W light bulb is converted to visible radiation.What is the average intensity of visible radiation at distance of 10 m.

0%
0.4 W / $m^2$
0%
0.04 W / $m^2$
0%
0.004 W / $m^2$
0%
0.0004 W / $m^2$

Explanation

Given:- Power of light bulb=

${ P }_{ o }= 100W$

Powerof visible radiation=

$P=5%$ of

${ P }_{ o }$

$P=\dfrac { 5 }{ 100 } \times 100$

$P=5W$

$Distance=d=10m$

solution:-By considering light bulb as a point source.

Since radiation trable by spherical wave fronts.

Intensity of visible radiation at a distance

$d=10m$

is given by :-

$I=\cfrac { P }{ 4\Pi { d }^{ 2 } } =\cfrac { 5 }{ 4\times 3.14\times { 10 }^{ 2 } }$

$I=0.00398{ W }/{ { m }^{ 2 } }$

$I\approx 0.004{ W }/{ { m }^{ 2 } }$

Choose the correct answer from the alternatives given.
The photon energy in units of

$eV$ for electromagnetic waves of wavelength

$2\ cm$ is then

0%
$2.5 \, \times \, 10^{-19}$
0%
$5.2 \, \times \, 10^{16}$
0%
$3.2 \, \times \, 10^{-16}$
0%
$6.2 \, \times \, 10^{-5}$

Explanation

Given: The wavelength of the electromagnetic wave is

$2\ cm$ .

To find: The photon energy in terms of

$eV$ .

The energy of the photon of particular wavelength is given by:

$E \, = \, \dfrac{hc}{\lambda}\\= \, \dfrac{6.6 \, \times \, 10^{-34} \,\times \, 3 \, \times \, 10^{8}}{2 \, \times \, 10^{-2}} \\\Rightarrow 9.9 \, \times \, 10^{-24} \,J$

Convert the energy into

$eV$ .

$E\ (eV)= \, \dfrac{9.9 \, \times \, 10^{-24}}{1.6 \, \times \, 10^{-19}} \, eV \, = \, 6.2 \, \times \, 10^{-5} \, eV$

Option

$(D)$ is correct.

$100 W$ sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is

$589 nm$ . The number of photons delivered per second to the sphere is

0%
$3 \, \times \, 10^{15}$
0%
$3 \, \times \, 10^{10}$
0%
$3 \, \times \, 10^{20}$
0%
$3 \, \times \, 10^{19}$

Explanation

Here,

$\lambda = 589 \times 10^{-9}m, h = 6.63 \times 10^{-34}J s$

$P = 100 W$
Energy of a photon,

${E} = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34}\times 3 \times 10^8}{589 \times 10^{-9}}$

Number of photons deliverved per second

$n = \dfrac{P}{E} = \dfrac{100}{3.38 \times 10^{-19}} = 3\times 10^{20}$

A blue lamp mainly emits light of wavelength 4500

$\mathring{A}$ . The lamp is rated at 150 W and 8% of the energy is emitted as visible light. The number of photons emitted by the lamp per second is:

0%
$3 \, \times \, 10^{19}$
0%
$3 \, \times \, 10^{24}$
0%
$3 \, \times \, 10^{20}$
0%
$3 \, \times \, 10^{18}$

Explanation

Number of photons emitted per second

$= \dfrac{8% of P}{E} = \dfrac{8P\lambda}{100hc}= \dfrac{8\times 150\times 4500\times 10^{-10}}{100\times 6.63 \times 10^{-34}\times 3 \times 10^8}$

$=3 \times 10^{19}$ photons per second

Which of the following functions is performed by a photo-electric cell?

0%
It converts electrical energy into light
0%
It converts light energy into electrical energy
0%
It conserves sound energy
0%
It converts electrical energy into sound

If h is Plancks constant, the momentum of a photon of wavelength 0.01

$A^o$ is

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

Explanation

Momentum of photo,

$p = \dfrac{E}{c} = \dfrac{h\nu}{c}$ where E is the energy of a photon and c is the velocity of light.

$\therefore$

$p = \dfrac{hc}{c\lambda}$

$\,\,\,\,$

$\left[\because \nu=\dfrac{c}{\lambda}\right]$

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

A laser beam is for locating distant objects because

0%
it is monochromatic
0%
it is constant
0%
it is not observed
0%
it has small angular spread

The maximum value of photoelectric current is called:

0%
base current
0%
Saturation current
0%
collector current
0%
emitter current

The energy flux of sunlight reaching the surface of the earth is

$1.388 \, \times \, 10^3 Wm^{-2}$ . The photons in the sunlight have an average wavelength of 550 nm. How many photons per square metre are incident on the earth per second?

0%
$4 \, \times \, 10^{21}$
0%
$4 \, \times \, 10^{34}$
0%
$4 \, \times \, 10^{31}$
0%
$4 \, \times \, 10^{28}$

Explanation

Here,

$I = 1.388 \times 10^3W m^{-2}$

$\lambda = 550 \times 10^{-9}m, h = 6.63 \times 10^{-34}J s$

Number of photos incident on earth's surface per second per square metre is

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

$\,\,\,\,$

$\left[\therefore E = \dfrac{hc}{\lambda}\right]$

$= \dfrac{1.388 \times 10^3 \times 550 \times 10^{-9}}{6.63 \times 10^{-34}\times 3\times 10^{-8}} = 4 \times 10^{21}$

The wavelength of light in the visible region is about

$390 nm$ for violet colour and about

$760 nm$ for red colour. The energy of photon in

$eV$ at violet end is

0%
$2.32$
0%
$3.19$
0%
$1.42$
0%
$4.13$

Explanation

Energy of the incident photon,

$E = h\nu - \dfrac{hc}{\lambda}$
For violet light,

$\lambda= 390 nm$

$\therefore$ Incident photon energy,

$E = \dfrac{6.63 \times 10^{-34} \times 3 \times {10^8}}{390\times 10^{-9}}$ J

$=5.1 \times 10{-19}J = \dfrac {5.1 \times 10^{-19}}{{1.6} \times 10^{19}}eV = 3.19 eV$

The linear momentum of a 3 MeV photon is:

0%
0.01 $eV \, sm^{-1}$
0%
0.02 $eV \, sm^{-1}$
0%
0.03 $eV \, sm^{-1}$
0%
0.04 $eV \, sm^{-1}$

Explanation

Energy of photon,

$E = 3MeV = 3 \times 10^6 eV$

Linear momentum of the photon,

$p =\dfrac{E}{c}$
where c is the speed of light in vacuum

$p = \dfrac{3 \times 10^6 eV}{3\times 10^8 m s^{-1}} = 10^{-2}eV s m^{-1} = 0.01 eV s m^{-1}$

A proton, a neutron, an electron and an

$\alpha$ -particle have same energy. Then their de Broglie wavelengths compare as :

0%
$\lambda_p\, = \,\lambda_n\, > \,\lambda_e \,> \,\lambda_\alpha$
0%
$\lambda_\alpha\, < \,\lambda_p\, = \,\lambda_n \,< \,\lambda_e$
0%
$\lambda_e\, < \,\lambda_p\, = \,\lambda_n \,> \,\lambda_\alpha$
0%
$\lambda_e\, = \,\lambda_p\, = \,\lambda_n \, = \,\lambda_\alpha$

Explanation

Kinetic energy of particle,

$K = \dfrac{1}{2}mv^2 \,\,\,\, or \,\,\,\,\, mv = \sqrt{2mK}$

de Broglie wavelength,

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

For the given value of

$K, \lambda \alpha \dfrac{1}{\sqrt{m}}$

$\therefore \lambda_p : \lambda_n : \lambda_e : \lambda_{\alpha} = \dfrac{1}{\sqrt{m_p}} : \dfrac{1}{\sqrt{m_n}}: \dfrac{1}{\sqrt{m_e}} :\dfrac{1}{\sqrt{m_\alpha}}$

Since

$m_p = m_n, \,\, hence\,\, \lambda_p = \lambda_n$

$As\ m_{\alpha} > m_p, \,\, therefore \,\, \lambda_{alpha} < \lambda_p$

$m_e < m_n, therefore \lambda_n < \lambda_n$

Hence

$\lambda_{alpha} < \lambda_p = \lambda_n < \lambda_e$

When the velocity of an electron increases, its de Broglie wavelength:

0%
increases
0%
decreases
0%
remains same
0%
may increase or decrease

Explanation

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

$\lambda _D \propto \dfrac{1}{v}$
velocity of an electron increases, its de Broglie wavelength decreases.

Option B

The de Broglie wavelength of an electron in a metal at

$27^{\circ}C$ is

$(Given \, m_e \, = \, 9.1 \, \times \, 10^{-31} \, kg,\, k_B \, = \, 1.38 \, \times \, 10^{-23} \, J \, K^{-1})$

0%
$6.2 \, \times \, 10^{-9} \, m$
0%
$6.2 \, \times \, 10^{-10} \, m$
0%
$6.2 \, \times \, 10^{-8} \, m$
0%
$6.2 \, \times \, 10^{-7} \, m$

Explanation

Here, T = 27 + 273 = 300 K
For an electron in a metal, momentum

$p = \sqrt{3mk_{B}T}$
de Broglie wavelength of an electron is

$\lambda \, = \dfrac{h}{p} \, = \dfrac{h}{\sqrt{3mk_{B}T}}$

$= \dfrac{h}{\sqrt{3 \times \left ( 9.1 \times 10^{-31} \right ) \times \left ( 1.38 \times 10^{-23} \right )\times 300}}$

$= \, 6.2 \times 10^{-9} m$

A monochromatic light of frequency

$3 \, \times \, 10^{14} \, Hz$ is produced by a LASER, emits the power of

$3 \, \times \, 10^{-3}\, W$ . Find how many numbers of photons are emitted per second.

0%
$1.5 \, \times \, 10^{16}$
0%
$2.5 \, \times \, 10^{16}$
0%
$4.5 \, \times \, 10^{16}$
0%
$8.5 \, \times \, 10^{16}$

Explanation

Energy of a photon,

$E = h\nu$

$E = (6.63 \times 10^{-34}J s)(3\times 10^{14}Hz) = 19.89 \times 10^{-20}J$
Power of the source,

$P = 3\times 10^{-3}W$
Number of photons emitted per second,

$N = \dfrac{P}{E} = \dfrac{3 \times 10^{-3}W}{19.89 \times 10^{-20}J }= 1.5 \times 10^{16}$

A particle of mass

$4m$ at rest decays into two particles of masses

$m$ and

$3m$ having non-zero velocities. The ratio of the de Broglie wavelengths of the particles 1 and 2 is:

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

Explanation

According to law of conservation of linear momentum, two particles will have equal and opposite momentum.
The de Broglie wavelength is given by

$\lambda = \dfrac{h}{p} \, \, \, \, \therefore \dfrac{\lambda_1}{\lambda_2} = 1$

There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light of 500 nm. The ratio of number of photons of X-rays to the photons of visible light of the given wavelength is:

0%
1:500
0%
1:400
0%
1:300
0%
1:200

Explanation

Here,

$P = 100 W,\lambda_1 = 1 nm, \lambda_2= 500 nm$

Let

$n_1$ and

$n_2$ be the number of photons of X-rays and visible light emitted from the two sources

$\therefore$

$n_1\dfrac{hc}{\lambda_1} = n_2 \dfrac{hc}{\lambda_2}$ or

$\dfrac{n_1}{\lambda_1}= \dfrac{n_2}{\lambda_2}$ n or

$\dfrac{n_1}{n_2} = \dfrac{\lambda_2}{\lambda_2} =\dfrac{1}{500}$

Relativistic corrections become necessary when the expression for the kinetic energy

$\dfrac{1}{2} mv^2$ , becomes comparable with

$mc^2$ , where m is the mass of the particle. At what de Broglie waxelength will relativistic corrections become important for an electron?

0%
$\lambda \, = \, l \, nm$
0%
$\lambda \, = \, l0 \, nm$
0%
$\lambda \, = \, l0^{-1} \, nm$
0%
$\lambda \, = \, l0^{-4} \, nm$

Explanation

We know that the de Broglie's equation is written as:

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

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

velocity of electrn,

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

The value of plank's constant is

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

and the mass of the electron is

$m = 9 \times 10^{-31}kg$

$(a)$

For

$\lambda_1 = 10nm = 10^{-8}m$

$v_1 = \dfrac{6.6\times 10^{-34}}{(9\times 10^{-31})\times 10^{-8}}$

$= \dfrac{2.2}{3} \times 10^5 \approx 10^5m/s$

$(b)$

For

$\lambda_2 = 10^{-1}nm = 10^{-10}m$

$v_2 = \dfrac{6.6\times 10^{-34}}{9\times 10^{-31} \times 10^{-10}}$

$\approx 10^7 m/s$

$(c)$

For

$\lambda_3 = 10^{-4}nm = 10^{-13}m$

$v_3 = \dfrac{6.6\times 10^{-34}}{9\times 10^{-31} \times 10^{-13}}$

$\approx 10^{10} m/s$

$(d)$

For

$\lambda_4 = 10^{-6}nm = 10^{-15}m$

$v_4 = \dfrac{6.6\times 10^{-34}}{9\times 10^{-31} \times 10^{-15}}$

$\approx 10^{12} m/s$

Since

$v_3$ and

$v_4$ are greater than velocity of light . So, the relativistic correction can occur in these speeds of the electron.

Thus, option (C) and (d) are required wavelengths.

Consider the four gases hydrogen, oxygen, nitrogen and helium at the same temperature. Arrange them in the increasing order of the de Broglie wavelengths of their molecules.

0%
Hydrogen, helium, nitrogen, oxygen
0%
Oxygen, nitrogen, hydrogen, helium
0%
Oxygen, nitrogen, helium, hydrogen
0%
Nitrogen, oxygen, helium, hydrogen

Explanation

de Broglie wavelength of a gas molecule

$\lambda = \dfrac{h}{\sqrt{3mk_BT}}$
where, T = Absolute temperature
kg = Boltzmann's constant
m = Mass of gas molecule
For the same temperature

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

$M_0 > M_N > M_{He} > M_{H2}$

$\therefore \lambda_O < \lambda_N < \lambda_{He} < \lambda_{H2}$

Two particles

$A_1$ and

$A_2$ of masses

$m_1, \, m_2 \, (m_1 \, > \, m_2)$ have the same de Broglie wavelength. Then

0%
their momenta are the same.
0%
their energies are the same.
0%
momentum of $A_1$ is less than momentum of $A_2$
0%
energy of $A_1$ is more than the energy of $A_2$

Explanation

$\lambda = \dfrac{h}{p} \, or \, p = \dfrac{h}{\lambda} \, or \, p \propto \dfrac{1}{\lambda}$

$\therefore \dfrac{p_1}{p_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{\lambda}{\lambda} = 1 \, or \, p_1 = p_2$ ..............(1)

Also

$E = \dfrac{1}{2} \dfrac{p^2}{m} = \dfrac{1}{2m} \dfrac{h^2}{\lambda^2} \, \, \, \, \left(\therefore p = \dfrac{h}{\lambda}\right)$

$E \propto \dfrac{1}{m} \, \, \therefore \dfrac{E_1}{E_2} = \dfrac{m_2}{m_1} \, < \, 1 \, or \, E_1 \, < \, E_2$ ................(2)

From above equation: Option A

A particle

$A$ of mass

$m$ and initial velocity

$v$ collides with a particle

$B$ of mass

$\cfrac{m}{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths

${ \lambda }_{ A }$ and

${ \lambda }_{ B }$ after the collision is :

0%
$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 1 }{ 2 }$
0%
$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 1 }{ 3 }$
0%
$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =2$
0%
$\cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { 2 }{ 3 }$

Explanation

$mv=m{ v }_{ A }+\cfrac { m }{ 2 } { v }_{ B }$ (conservation of linear momentum)

$\quad \therefore v={ v }_{ A }+\cfrac { { v }_{ B } }{ 2 } ={ v }_{ B }-{ v }_{ A }\quad$ (elastic collision)

$\therefore \cfrac { { v }_{ B } }{ { v }_{ A } } =4\quad \quad \therefore \cfrac { { \lambda }_{ A } }{ { \lambda }_{ B } } =\cfrac { { { m }_{ B }v }_{ B } }{ { { m }_{ A }v }_{ A } } =2$

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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