MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 14

The ratio of the deBroglie wave length for the electron and proton moving with same velocity is:

[

$m_p$ - mass of propton,

$m_e$ -mass of electron]

0%
$m_p:m_e$
0%
$m_p^2:m_e^2$
0%
$m_e:m_p$
0%
$m_e^2:m_p^2$

Explanation

$\dfrac{\lambda_e}{\lambda_p}=\dfrac{\dfrac{h}{m_eV}}{\dfrac{h}{m_pV}}$

$=\dfrac{m_p}{m_e}$

The de-Broglie wavelength of a free electron with kinetic energy

$'E'$ is

$\displaystyle \lambda$ . If the kinetic energy of the electron is doubled, the de-Broglie wavelength is:

0%
$\displaystyle \frac { \lambda }{ \sqrt { 2 } }$
0%
$\displaystyle \sqrt { 2 } \lambda$
0%
$\displaystyle \frac { \lambda }{ 2 }$
0%
$\displaystyle 2\lambda$

Explanation

Kinetic energy initially is given by:

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

$\implies p=\sqrt {2m E}$

where

$m:$ Mass of the particle

$v:$ Speed of the particle

$p:$ Momentum of the particle

De-broglie wavelength is given by,

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

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

Given

$E_2=2E_1$

$\implies \lambda_2=\dfrac{\lambda_1} {\sqrt 2}$

Given

$\lambda_1=\lambda, \quad \therefore \lambda_2=\dfrac{\lambda}{\sqrt 2}$

An atom of mass M which is in the state of rest, emits a photon of wavelength

$\lambda$ . As a result atom will deflect with the kinetic energy equation to (h is planck's constant).

0%
$\displaystyle\frac{h^2}{M\lambda^2}$
0%
$\displaystyle\frac{1}{2}\frac{h^2}{M\lambda^2}$
0%
$\displaystyle\frac{h}{M\lambda}$
0%
$\displaystyle\frac{1}{2}\frac{h}{M\lambda}$

Explanation

Momentum of photon,

$p=\displaystyle\frac{h}{\lambda}$
Kinetic energy of photon of mass M, K

$=\displaystyle\frac{P^2}{2M\lambda}$
or

$K=\displaystyle\frac{h^2}{2M\lambda^2}$ .

What is the ratio of wavelength of a photon and that of an electron of mass, m of the same energy

$E$ ?

0%
$C\sqrt {\dfrac {2m}{E}}$
0%
$\sqrt {\dfrac {2m}{E}}$
0%
$\dfrac{1}{C}\sqrt {\dfrac {2m}{E}}$
0%
$\sqrt {\dfrac {m}{E}}$

Explanation

Wavelength of the photon is related to its energy by:

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

For an electron its deborglie wavelength is given by:

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

Squaring,

$\lambda_e^2 = \cfrac{h^2}{m^2v^2}$

$= \cfrac{1/2h^2}{Em}$

$\lambda_e = \cfrac{h}{\sqrt{2Em}}$

$\cfrac{\lambda_p}{\lambda_e} = c \sqrt{ \cfrac{2m}{E} }$

In hydrogen spectrum, the wavelength of the line is 656 nm, where as in the spectrum of a distant galaxy, the line wavelength is 706 nm. Estimated speed of galaxy with respect to the earth is :

0%
$2 \times 10^8 m/s$
0%
$2 \times 10^7 m/s$
0%
$2 \times 10^6 m/s$
0%
$2 \times 10^5 m/s$

Explanation

$\dfrac { \triangle \lambda}{\lambda} = \pm \dfrac {v}{c}$

Here (-) sign to be used for appraoching (+) sign to be used for receding.

$\dfrac { \triangle \lambda}{\lambda} = \dfrac {706-656}{656} = \pm = v/c$

$\dfrac {v}{c} = \dfrac {50}{656}$

$v = \dfrac {50}{656} \times c$

$v = \dfrac {50}{656} \times 3 \times 10^8 = 2 \times10^7 m/s$

Find the force exerted by the light beam on the surface.

0%
$6.66\times 10^{8}N$
0%
$6.66\times 10^{-8}N$
0%
$3.33\times 10^{8}N$
0%
$3.33\times 10^{-8}N$

Explanation

Power of the beam

$P =10 \ W$
Wavelength

$\lambda = 500 \ nm = 500\times 10^{-9} \ m$
Energy of photon

$E = \dfrac{hc}{\lambda}$
Number of photons absorbed per second

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

$\therefore$

$N = \dfrac{10\times 500\times 10^{-9}}{6.62\times 10^{-34}\times 3\times 10^8} = 2.52\times 10^{19}$
Force exerted by the light beam

$F = \dfrac{Nh}{\lambda}$

$\therefore$

$F = \dfrac{(2.52\times 10^{19})(6.62\times 10^{-34})}{500\times 10^{-9}} = 3.33\times 10^{-8 }$ N

A small piece of

$Cs$ (work function

$=1.9eV$ ) is placed 22 cm away from a large metal plate. The surface charge density on the metal plate is

$1.21\times 10^{-9}Cm^{-2}$ . Now, light of

$460nm$ wavelength is incident on the piece of

$Cs$ . Find the maximum and minimum energies of photo electrons on reaching the plate. Assume that no change occurs in electric field produced by the plate due to the piece of

$Cs$ .

0%
Minimum energy $=20.83eV$ ,
Miximum energy $=32.63eV$ .
0%
Minimum energy $=25.83eV$ ,
Miximum energy $=35.63eV$ .
0%
Minimum energy $=21.83eV$ ,
Miximum energy $=30.63eV$ .
0%
Minimum energy $=29.83eV$ ,
Miximum energy $=30.63eV$ .

A particular of mass M at rest decays into two particles of masses

${ m }_{ 1 }$ and

${ m }_{ 2 }$ having non-zero velocities. The ratio of debroglie wavelength of paticle is

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

An electron of mass 'm' has de-Broglie wavelength '

$\lambda$ ' when accelerated through potential difference 'V'. When proton of mass 'M', is accelerated through potential difference

$9V$ , the de-Broglie wavelength associated with it will be : (Assume that wavelength is determined at low voltage)

0%
$\dfrac{\lambda}{3} \sqrt{\dfrac{M}{m}}$
0%
$\dfrac{\lambda}{3} . \dfrac{M}{m}$
0%
$\dfrac{\lambda}{3} \sqrt{\dfrac{m}{M}}$
0%
$\dfrac{\lambda}{3} . \dfrac{m}{M}$

Explanation

De Broglie wavelength

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

where

$K$ is the kinetic energy.

Kinetic energy

$K =eV$

We get

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

$\implies$

$\lambda \propto \dfrac{1}{\sqrt{mV}}$ ........(1)

Given :

$V_1 = V$

$m_1 = m$

$m_2 = M$

$V_2 = 9V$

From equation (1), we get

$\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{m_1 V_1}{m_2V_2}}$

$\dfrac{\lambda_2}{\lambda} = \sqrt{\dfrac{m V}{M(9V)}}$

$\implies$

$\lambda_2 =\dfrac{\lambda}{3} \sqrt{\dfrac{m }{M}}$

Find the number of photons absorbed per second by the surface.

0%
$2.52\times 10^{18}$
0%
$2.52\times 10^{19}$
0%
$2.52\times 10^{-18}$
0%
$2.52\times 10^{-19}$

Explanation

Power of the beam

$P =10 \ W$
Wavelength

$\lambda = 500 \ nm = 500\times 10^{-9} \ m$
Energy of photon

$E = \dfrac{hc}{\lambda}$
Number of photons absorbed per second

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

$\therefore$

$N = \dfrac{10\times 500\times 10^{-9}}{6.62\times 10^{-34}\times 3\times 10^8} = 2.52\times 10^{19}$

At what rate, photons are emitted from the lamp?

0%
$3\times 10^{20} photons s^{-1}$
0%
$2\times 10^{20} photons s^{-1}$
0%
$5\times 10^{20} photons s^{-1}$
0%
$1\times 10^{20} photons s^{-1}$

Explanation

Power of the beam

$P =100 \ W$
Wavelength

$\lambda = 5890 \ A^o = 5890\times 10^{-10} \ m$
Energy of photon

$E = \dfrac{hc}{\lambda}$
Number of photons absorbed per second

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

$\therefore$

$N = \dfrac{100\times 5890\times 10^{-10}}{6.62\times 10^{-34}\times 3\times 10^8} = 3\times 10^{20}$ photons per second

A particle A with a mass

$m_A$ is moving with a velocity v and hits a particle B of mass

$m_B$ at rest. If motion is one dimensional and take the collision is elastic, then the change in the de Broglie wavelength of the particle A is

0%
$\dfrac{h}{2m_A v} \left[\dfrac{(m_A \, + \, m_B)}{(m_A \, - \, m_B)} \, - \, 1\right]$
0%
$\dfrac{h}{m_A v} \left[\dfrac{(m_A \, - \, m_B)}{(m_A \, + \, m_B)} \, - \, 1\right]$
0%
$\dfrac{h}{m_A v} \left[\dfrac{(m_A \, + \, m_B)}{(m_A \, - \, m_B)} \, - \, 1\right]$
0%
$\dfrac{2h}{m_A v} \left[\dfrac{(m_A \, + \, m_B)}{(m_A \, - \, m_B)} \, + \, 1\right]$

Explanation

According to law of conversation of momentum

$m_Av \, + \, m_B \, \times \, 0 \, = \, m_Av_A \, + \, m_Bv_B$

$m_A(v - v_B) \, = \, m_Bv_B$ ....(i)

According to law of conversation of klinetic energy

$\dfrac{1}{2}m_Av^2 \, = \, \dfrac{1}{2}m_Av^2_A \, + \, \dfrac{1}{2}m_Bv^2_B$

$m_A(v^2 - V^2_A) = m_Bv^2_B$

$m_A (v - v_A)(v + v_A) \, = \, m_Bv^2_b$ ...(ii)

Dividing (ii) by (i), we get

$v + v_A = v_B \, or \, v = v_B - v_A$ ...(iii)

Solving (i) and (iii), we get

$v_A = \left ( \dfrac{m_A - m_B}{m_A + m_B} \right )v \,$ and

$\, v_B = \left ( \dfrac{2m_A}{m_A + m_B} \right )v$

$\lambda_{initial} = \dfrac{h}{m_Av}$ and

$\, \, \lambda_{final} = \dfrac{h}{m_Av_A} = \dfrac{h(m_A + m_B)}{m_A(m_A - m_B)v}$

$\therefore \Delta\lambda = \lambda_{final} - \lambda_{initial} = \dfrac{h}{m_Av}\left [ \dfrac{(m_A + m_B)}{(m_A - m_B)} -1\right ]$

An electromagnetic wave of wavelength

$\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength

$\lambda'$ , then

0%
$\lambda \, = \, \dfrac{mc}{h} \, \lambda'^2$
0%
$\lambda \, = \, \dfrac{3mc}{2h} \, \lambda'^2$
0%
$\lambda \, = \, \dfrac{2mc}{h} \, \lambda'^2$
0%
$\lambda \, = \, \dfrac{5mc}{h} \, \lambda'^2$

Explanation

Kinetic energy of emitted electron = Energy of incident photon
i.e.

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

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

$\left[ \because mv = p, \nu =\dfrac{c}{\lambda}\right]$
or

$p = \sqrt{\dfrac{2mhc}{\lambda}}$

de-Broglie wavelength of emitted electrons

$\lambda' = \dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2mhc}{\lambda}}}$ or

$\lambda' = \sqrt{\dfrac{h\lambda}{2mc}}$

$\therefore \lambda = \dfrac{2mc}{h}\lambda'^2$

$100\ W$ sodium lamp is radiating light of wavelength

$5890 \overset {o}{A}$ , uniformly in all directions. At what distance from the lamp, the average flux is 1 photon

$(cm^2-s)^{-1}$ ?

0%
48600 km
0%
48860 km
0%
50000 km
0%
58900 km

Explanation

Power of the beam

$P =100 \ W$

Wavelength

$\lambda = 5890 \ A^o = 5890\times 10^{-10} \ m$

Energy of photon

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

Number of photons absorbed per second

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

$\therefore$

$N = \dfrac{100\times 5890\times 10^{-10}}{6.62\times 10^{-34}\times 3\times 10^8} = 3\times 10^{20}$ photons per second

Photon flux at a distance

$r$ is

$\phi_{flux} = \dfrac{N}{4\pi r^2}$

where

$\phi_{flux} = 1 \ photon \ /cm^2/s = 10^4 \ photon /m^2/s$

$\therefore$

$10^4 =\dfrac{3\times 10^{20}}{4\pi r^2}$

We get

$r^2 = 0.239\times 10^{16}$

$\implies \ r = 0.48860 \times 10^8 \ m = 48860 \ km$

Two metallic plates

$A$ and

$B$ , each of area

$5 \, \times \, 10^{-4} \, m^2$ , are placed parallel to each other at a separation of

$1\ cm$ . Plate

$B$ carries a positive charge of

$33.7 \ pc$ . A monochromatic beam of light, with photons of energy

$5\ eV$ each, starts falling on plate

$A$ at

$t = 0$ so that

$10^{16}$ photons fall on it per square metre per second. Assume that one photoelectron is emitted for every

$10^6$ incident photons. Also assume that all the emitted photoelectrons are collected by plate

$B$ and the work function of plate A remains constant at the value

$2\ eV$ . Determine the number of photoelectrons emitted upto

$t = 10\ s$ .

0%
$2 \, \times \, 10^3\ N/C$
0%
$10^3 \ N/C$
0%
$5 \, \times \, 10^3\ N/C$
0%
$Zero$

Explanation

Area of each plate =

$5 \times 10^{-4} \, m^2$

Separation between plates =

$1 \, cm \, = \, 10^{-2} \, m$

$\dfrac{photons \, incident}{area \times time} = 10^{16}, \, t = 10 \, s$

$\therefore No.\, of \, photons \, incident \, =\, 10^{16} \times \, area \, \times \, time \, = \, 10^{16} \times (5 \times 10^{-4}) \times 10 = 5 \times 10^{13}$

$\therefore \, Required \, number \, of \, photoelectrons \, emitted \, = \, \dfrac{5 \times 10^{13}}{10^6}$

Number of photoelectrons =

$5 \times 10^7$

A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is

$1.813 \, \times \, 10^{-4}$ . The mass of the particle is

$(m_e \, = \, 9.1 \, \times \, 10^{-31} \, kg)$

0%
$1.67 \, \times \, 10^{-27} \, kg$
0%
$1.67 \, \times \, 10^{-37} \, kg$
0%
$1.67 \, \times \, 10^{-19} \, kg$
0%
$1.67 \, \times \, 10^{-14} \, kg$

Explanation

Given :

$\dfrac{v}{v_e} = 3$ and

$\dfrac{\lambda}{\lambda_e} = 1.813 \times 10^{-4}$

De Broglie wavelength of a moving particle having mass m and velocity v is given by

$\because P=mv=\dfrac{h}{\lambda_D}$

$h=mv \lambda _D=m_e v_e \lambda _{D_e}=mv \lambda _D$

Mass of the particle,

$m = m_e \left[\dfrac{\lambda_{D_e}}{\lambda}\right]\left[\dfrac{v_e}{v}\right]$

Substituting the value, we get

$m = 9.1 \times 10^{-3} \times \dfrac{1}{1.813 \times 10^{-4}}\times \dfrac{1}{3}$

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

The intensity of radiation from a human body is maximum around a certain wavelength. A photon of this wavelength can just excite an electron from the valence to the conduction band of a semiconductor used in a night vision device. Assume that the black body radiation law holds for the human body. The band gap of such a semiconductor is close to

0%
$0.1 eV$
0%
$0.5 eV$
0%
$1.0 eV$
0%
$2.0 eV$

Explanation

We know that:-

$\lambda_mT=b$ , where

$\lambda_m$ is the wavelength of maximum intensity

$T= temperature$ and

$b=Stefan's \hspace{1mm}constant$

$=2.88\times 10^{-3}$

Here,

$T=$ body temperature

$37^{o}C=310K$

Hence,

$\lambda_m=\dfrac{b}{T}=\dfrac{2.88\times 10^{-3}}{310}m=9.3\times 10^{-6}m$

Now, energy of photon,

$E=\dfrac{hc}{\lambda_m}=\dfrac{6.63\times10^{-34}\times 3\times 10^8}{9.3\times 10^{-6}}J$

$E=2.14\times10^{-20}J=\dfrac{2.14\times10^{-20}}{1.6\times10^{-19}}eV$

$\implies E \approx0.1eV=Band \ gap \ of \ semi-conductor$

Hence, answer is option-(A).

In a Young double slit experiment, the distance between the slits is 50

$\mu m$ and the distance of the screen from the slits is 5 cm. An infrared monochromatic light is mood to produce an interference pattern with fringe width 1 mm. If the light source is replaced by an electron source to produce the same fringe width then the speed of the electrons is approximately

0%
$70 m/s$
0%
$700 m/s$
0%
$7000 m/s$
0%
$70000 m/s$

Explanation

We know that :-

$w=\dfrac{\lambda D}{d}$

Given :-

$d=50\times10^{-6}m,$

$w=5\times 10^{-2}m$ and

$w=10^{-3}m$

$\implies \lambda=\dfrac{wd}{D}=\dfrac{10^{-3}\times 50\times 10^{-6}}{5\times10^{-2}}m$

$\implies \lambda=10^{-6}m$

For an electron to produce same fringe width, its de-Broglie wavelength should be same as

$\lambda$ .

Now,for electron:-

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

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

$\implies v=\dfrac{6.63\times 10^{-34}}{9.1\times10^{-31}\times 10^{-6}}m/s$

$\implies v\approx700m/s$

Hence, answer is option-(B).

An electron (mass m) with an initial velocity

$\bar{v}\, = \,v_0\,\,\hat{i}(v_0 > 0)$ is in an electric field

$\bar{E} = -E_0\,\,\hat{i}$ (

$E_0\, = \,constant \,>\, 0$ ). Its de Broglie wavelength at time t is given by

Assume:

$\left ( \, \lambda_0 = \dfrac{h}{mv_0} \right )$

0%
$\frac{\lambda_0}{\left ( 1\,+\,\dfrac{eE_0t}{mv_0} \right )}$
0%
$\lambda_0\left ( 1\,+\,\dfrac{eE_0t}{mv_0}\right )$
0%
$\lambda_0$
0%
$\lambda_0t$

Explanation

Here,

$\bar{E} = -E_0\, \hat{i}$ ; initial velocity

$\bar{v} = v_0 \,\, \hat{i}$ Force action on electron due to electric field

$\bar{F} = (-e)(-E_0 \, \hat{i}) = eE_0\, \hat{i}$
Acceleration produced in the electron,

$\bar{a} = \dfrac{\bar{F}}{m} = \dfrac{eE_0}{m} \, \hat{i}$
Now, velocity of electron after time t,

$\bar{v}_t = \bar{v} + \bar{a} \, t = \left ( v_0 + \dfrac{eE_0}{m} \right )\, \hat{i} \,\,\, or \,\,\, \left | \bar{v}_t \right |= v_0 + \dfrac{eE_0t}{m}$

Now,

$\lambda_t = \dfrac{h}{mv_t}= \dfrac{h}{m\left ( v_0 + \dfrac{eE_0t}{m} \right )} = \dfrac{h}{mv_0\left ( 1 + \dfrac{eE_0t}{mv_0} \right )}$

$\lambda_t = \dfrac{\lambda_0}{\left ( 1 +\dfrac{eE_0t}{mv_0} \right )}$

$\left ( \because \, \lambda_0 = \dfrac{h}{mv_0} \right )$

Assume that a molecule is moving with the root mean square speed at temperature 300 K. The de Broglie wavelength of nitrogen molecule is (Atomic mass of nitrogen = 14.0076 u, h =

$6.63 \, \times \, 10^{-27} \, J \, s, \, k_B \, = \, 1.38 \, \times \, 10^{-23} J \, K^{-1}, \, 1 \, u \, = \, 1.66 \,. \times \, 10^{-27} \, kg)$

0%
$2.75 \, \times \, 10^{-11} \, m$
0%
$2.75 \, \times \, 10^{-12} \, m$
0%
$3.24 \, \times \, 10^{-11} \, m$
0%
$3.24 \, \times \, 10^{-12} \, m$

Explanation

Mean Kinetic energy if a molecule:

$\dfrac{1}{2}\, mv_{rms}^{2} \, = \, \dfrac{3}{2}k_{B}T$

Here, m =

$2 \times \, 14.0067 \, = \, 28.02 \, u$

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

$= \, \dfrac{6.63 \, \times 10^{-34}}{\sqrt{3\times \left ( 28.02 \times 1.66 \times 10^{-27} \right )} \times 1.38 \times 10^{-23}\times 300}$

$= \, 2.75 \times 10 ^{-11}m$

A beam of light has three wavelengths

$4144\overset {\circ}{A}, 4972 \overset {\circ}{A}$ and

$6216\overset {\circ}{A}$ with a total intensity of

$3.6\times 10^{-3} W/m^{2}$ equally distributed amongst the three wavelengths. The beam falls normally on an area

$1.0\ cm^{2}$ of a clean metallic surface of work function

$2.3\ eV$ . Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.

0%
$11\times 10^{11}$ .
0%
$12\times 10^{11}$ .
0%
$13\times 10^{11}$ .
0%
$21\times 10^{11}$ .

Explanation

Threshold wavelength,

$\lambda_0=\dfrac{hc}{\phi}$

$\lambda_0=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{2.3\times (1.6\times 10^{-19})}=5.404\times 10^{-7}$

$=5404\mathring { A }$

Photons with wavelength greater than

$5404\mathring{A}$ will not eject electrons

Energy=intensity of each

$\lambda \times$ Area of the

$=\dfrac{3.6\times 10^{-3}}{3}\times (1.0cm^2)=1.2\times 10^{-7}W$

$(t=1)$

$E=(1.2\times 10^{-7})(2)=2.4\times 10^{-7}J$

$(t=2)$

No. of photons

$n_1$ due to

$\lambda_1\,\,4144\mathring{A}$

$n_1=\dfrac{(2.4\times 10^{-7})(4144\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^8)}=0.5\times 10^{12}$

No. of photons

$n_2$ due to

$\lambda_2\,\,4972\mathring {A}$

$n_1=\dfrac{(2.4\times 10^{-7})(4972\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^8)}=0.575\times 10^{12}$

$N=n_1+n_2=(0.5+0.575)\times 10^{12}$

$=1.075\times 10^{12}$

$=10.75\times 10^{11}\approx 11\times 10^{11}$ (A)

A hydrogen -like atom is in a higher energy level of quantum numberThe excited atom make a transition to first excited state by emitting photons of total energy 27.2 eV. The atom from the same excited state by successively emitting two photons. If the energy of one photon is 4.25 eV, find the energy of other photon:

0%
$5.95 eV$
0%
$6.25 eV$
0%
$6.95 eV$
0%
$7.80 eV$

A particle of mass

$M$ at rest decays into two particle of masses

${m}_{1}$ and

${m}_{2}$ having non zero velocities. The ratio of the de Broglie wavelength of particles

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

0%
$\dfrac {{m}_{1}}{{m}_{2}}$
0%
$\dfrac {{m}_{2}}{{m}_{1}}$
0%
$1:1$
0%
$\sqrt { \dfrac { { { m }_{ 2 } } }{ { { m }_{ 1 } } } }$

Explanation

Let velocity of particle of mass

$m_1$ is

$v_1$ and velocity of particle of mass

$m_2$ is

$v_2$ .

There is no external force is acting so momentum will be conserved.

Before Decay, Linear momentum

$= 0$ ,

After decay, Linear momentum

$= m_1v_1+m_2v_2$

$\Rightarrow m_1v_1+ m_2v_2=0$

$\Rightarrow m_1v_1=-m_2v_2$

Magnitude

$m_1v_1=m_2v_2$

de Broglie wavelength,

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

$\Rightarrow \dfrac{\lambda_1}{\lambda_2}=\dfrac{m_2v_2}{m_1v_1}=1$

Hence,

$\lambda_1 : \lambda _2=1:1$

Consider a hypothetical hydrogen like atom. The wavelength in

$A$ for the spectral lines for transition from

$n=p$ to

$n=1$ are given by-

$\lambda =\dfrac { 1500\quad { p }^{ 2 } }{ { p }^{ 2 }-1 }$
Where

$p=2, 3, 4, .$ (given

$hc=12400\ eV/A$ )

0%
The wavelength of the least energetic and the most energetic photons in this series is $2000\ A$ , $1500\ A$
0%
Difference between energies of fourth and this orbit is $0.40\ eV$ .
0%
Energy of second orbit is $6.2\ eV$
0%
The ionization potential of this element is $8.27\ V$

Explanation

Given,

$\lambda=\dfrac{1500p^2}{p^2-1}=\dfrac{1500}{1-\dfrac{1}{p^2}}\ A^{\cdot}$

$\lambda_{max}$ at p=2 is

$2000 A^{\cdot}$

$\lambda_{min}$ at p=

$\infty$ is

$1500 A^{\cdot}$

$E_4=\ \dfrac{12400}{1500} \left( 1-\dfrac{1}{p^2}\right) =\dfrac{124}{15}\times \dfrac{15}{16}$

$E_3=\ \dfrac{124}{15} \left( 1-\dfrac{1}{9}\right)$

$= \dfrac{124}{15}\times \dfrac{8}{9}$

$E_2=\ \dfrac{124}{15} \left( 1-\dfrac{1}{4}\right)$

$=\dfrac{124}{15}\times\dfrac{3}{4} = \textbf{6.2 eV}$

Now,

$E_4 - E_3=\dfrac{124}{15}\times \left(\dfrac{15}{16}-\dfrac{8}{9}\right) = \dfrac{124}{15}\times\dfrac{7}{144}=\dfrac{868}{144\times 15}= \textbf{0.40 eV}$

and,

$E_{\infty} = \textrm{Ionisation Energy}$

$\textrm{Ionisation Energy}= -E_{\infty}=-\dfrac{124}{15}=-8.27\ eV$

$\textrm{Ionisation Potential}=\dfrac{\textrm{Ionisation energy}}{-e}$

$\textrm{Ionisation Potential}=8.27\ V$

Hence, Options

$\textbf{A,B,C,D}$ are correct.

The potential energy of particle of mass m varies as

$U (x) \, = \, \left\{\begin{matrix}E_0: \, \, \ 0 \, \leq \, x \, \leq \, 1 \\ 0: \, \ \, \, x \, > \, 1 \end{matrix}\right.$ The de Broglie wavelength of the particle in the range

$0 \, \leq \, x \, \leq \, 1 \, is \, \lambda_1$ and that in the range

$x \, > \, 1 \, is \, \lambda_2$ . If the total energy of the particle is

$2E_0$ , find

$\lambda_1,/\lambda_2.$

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

Explanation

de Broglie wavelength of a particle of ///

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

where K is the kinetic energy of the particle.

For

$0 \leq x \leq 1, \, U(x) = E_0$

As, total energy = Kinetic energy + potential energy

$\therefore 2E_0 = K_1 + E_0$

$K_1 = E_0$

$\therefore \lambda_1 = \dfrac{h}{\sqrt{2mK_1}} = \dfrac{h}{\sqrt{2mE_0}}$

For

$x > 1, U(x) = 0 \, \, \, \therefore K_2 = 2E_0$

$\therefore \lambda_2 = \dfrac{h}{\sqrt{2mK_2}} = \dfrac{h}{\sqrt{2m(2E_0)}}$

Divide (i) by (ii), we get

$\dfrac{\lambda_1}{\lambda_2} = \dfrac{h}{\sqrt{2mE_0}} \times \dfrac{\sqrt{2m(2E_0)}}{h} = \sqrt{2}$

After absorbing a slowly moving neutron of Mass

$m_N$ (momentum

$\approx 0$ ) a nucleus of mass M breaks into two nuclei of masses

$m_1$ and

$5m_1$

$(6m_1=M+m_N)$ respectively. If the de Broglie wavelength of he nucleus with mass

$m_1$ is

$\lambda$ , the de Broglie wavelength of the nucleus will be.

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

Explanation

Since initial momentum of the system is

$P_i=0$

Hence final momentum will also be

$P_f=0$ from conservation of momentum

Now, let momentum of two nuclei produced be

$P_1$ and

$P_2$ .

Then, both momentum will be opposite to each other and

$P_1=P_2$ in magnitude

Now,

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

Since,

$P_1=P-2\implies \lambda_1=\lambda_2=\lambda$

Answer-(C)

Root mean square velocity of the particle is v at pressure P. If Pressure is increased two times, then the r.m.s velocity becomes

0%
2v
0%
3v
0%
0.5 v
0%
v

What will be the number of photons emitted per second by a

$10 watt$ sodium vapour lamp assuming that

$90\%$ of the consumed energy is converted into light? Wavelength of sodium light is

$590nm;h=6.63\times 10^{-34}Js$

0%
$0.267\times 10^{18}$
0%
$0.267\times 10^{19}$
0%
$0.267\times 10^{20}$
0%
$0.267\times 10^{17}$

A surface irradiated with light of wavelength 480 nm gives out electrons with maximum velocity v m/s, the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity 2v m/s if it is irradiated by light of wavelength.

0%
325 nm
0%
360 nm
0%
384 nm
0%
300 nm

A laser used to weld detached retains emits light, with a wavelength

$652nm$ in pulses that are of

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

$0.6W$ . The energy in each pulse and in a singe photon are

0%
$7.5 \times 10^{15} eV , 2.7eV$
0%
$6.5 \times 10^{16} eV , 2.9eV$
0%
$6.5 \times 10^{16} eV , 2.7eV$
0%
$7.5 \times 10^{16} eV , 1.9eV$

Surface of sodium is illuminated by a light of

$6000$

$\dot{A}$ wavelength. Work function of sodium is

$1.6\ eV$ . Then minimum

$K.E.$ of emitted electrons is :

0%
$0\ eV$
0%
$1.53\ eV$
0%
$2.46\ eV$
0%
$4.14\ eV$

An electron is moving with velocity

$6.6 \times 10^3 mls.$ The DE-Imbroglio wavelength associated with electron is

$(mass of electron = 9 \times 10^{-31}kg,$ Plank's Constant =

$6.62 \times 10^{-34} J-S)$

0%
$1 \times 10^{-19} m$
0%
$1 \times 10^{-5} m$
0%
$1 \times 10^{-7} m$
0%
$1 \times 10^{-10} m$

Explanation

de broglie wavelength,

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

$\lambda = \dfrac{6.62 \times 10^{-34} }{g \times 10^{-31} \times 6.6 \times 10^3 m/s}$

$\lambda = 1 \times 10^{-7} m$

Hence (C) option is correct

A proton and an

$\alpha - particle$ accelerated through same voltage.The ratio of their de-broglie wavelength will be :

0%
$1 : 2$
0%
$2\sqrt 2 : 1$
0%
$\sqrt2 : 1$
0%
$2 : 1$

Explanation

$\lambda _\alpha = \lambda_P$

$\dfrac{h}{\sqrt{2m_\alpha q_\alpha V}} = \dfrac{h}{\sqrt{2m_P q_P V}}$

and we know

$m_\alpha = 4m_P$

$2q_\alpha = q_P$

$\dfrac{\lambda_P}{\lambda_a} = \dfrac{1}{2}$

Hence (A) option is correct

Sodium and copper have work functions 2.3 eV and 4.5 eV respectively. Then the ratio of the wavelengths . nearest to-

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

In a photoelectric experiment, the relation between applied potential difference between cathode and anode

$(V)$ and the photoelectric current

$(I)$ was found to be shown in graph below. If Planck's constant

$h = 6.6 \times 10^{-34}\ Js$ , the frequency of incident radiation would be nearly (in

$s^{-1}$ ).

0%
$0.436 \times 10^{18}$
0%
$0.436 \times 10^{17}$
0%
$0.775 \times 10^{15}$
0%
$0.775 \times 10^{16}$

Explanation

The kinetic energy is given as

${\rm{K = e}}{{\rm{V}}_{\rm{0}}}$

$K = 1.6 \times {10^{ - 19}} \times \left( { - 3.2} \right)$

$K = 5.12 \times {10^{ - 19}}$

By substituting the above value in

$5.12 \times {10^{ - 19}} = 6.6 \times {10^{34}} \times f + 3.2$

The frequency is $0.29 \times {1^{15}}$

The de-Broglie wavelength

$(\lambda_{B})$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state

$(\lambda_{G})$ by

0%
$\lambda_{B} = \lambda_{G}$
0%
$\lambda_{B} = 3\lambda_{G}$
0%
$\lambda_{B} = 3\lambda_{\dfrac {G}{3}}$
0%
$\lambda_{B} = 3\lambda_{\dfrac {G}{2}}$

A proton moves on a circular path of radius

$6.6\times 10^{-3}m$ in a perpendicular magnetic field of

$0.625\ tesla$ . The De broglie wavelength associated with the proton will be

0%
$1\overset {\circ}{A}$
0%
$0.1\overset {\circ}{A}$
0%
$0.01\overset {\circ}{A}$
0%
$0.001\overset {\circ}{A}$

If the momentum of an electron is changed by

$\Delta p,$ then the de-Brogile wavelength associated with it changes by

$0.05 \%$ . The initial momentum of electron will be :

0%
$\dfrac { \Delta p}{200}$
0%
$\dfrac { \Delta p}{199}$
0%
$199 \Delta p$
0%
$-2000 \Delta p$

Explanation

The initial momentum is given as,

$\dfrac{{d\lambda }}{\lambda } = - \dfrac{{dP}}{{{P_i}}}$

$\dfrac{{0.05}}{{100}} = - \dfrac{{\Delta P}}{{{P_i}}}$

${P_i} = - 2000\Delta P$

The initial momentum of electron is $- 2000\Delta P$ .

Two black objects of the same diameter, a sphere and a disc, are placed in front of a uniform beam of light. The plane of the disc is perpendicular to the light rays. The force acting upon them by the light is:

0%
zero
0%
bigger on the disc
0%
bigger on the sphere
0%
the same on both

An electron moving with velocity

$6.62 \times {10^3}$ . The De-Broglie wavelength associated with electron is (mass of electron

$= 9 \times {10^{ - 31}}Kg$
Plank's constant

$= 6.62 \times {10^{ - 34}}J - S$

0%
$1 \times {10^{ - 29}}m$
0%
$1 \times {10^{ - 5}}m$
0%
$1 \times {10^{ -7}}m$
0%
$1 \times {10^{ 10}}m$

The curve drawn between the maximum speed of photoelectron (v) and frequency

$(\upsilon)$ in photoelectric effect will be-

In an experiment, electrons are made to pass through a narrow slit of width

$d$ comparable to thier de-Broglie wavelength. They are detected on a screen at a distance

$D$ from the slit (see figure).
Which of the following graph can be expected to represent the number of electrons.

$N$ detected as function of the detected position

$y$ (

$y=0$ correxponds to the middle of the slit):

A source of power

$600kW$ emits photon which incident on a surface

$2.5m$ away out of which

$50\%$ is reflected back. What is the radiation pressure on the surface :

0%
$3.9\times10^{-5}Pa$
0%
$7.8\times10^{-5}Pa$
0%
$11.7\times10^{-5}Pa$
0%
$15.6\times10^{-5}Pa$

Explanation

Given,

$p=600kW=6\times 10^5 W$

$r=2.5m$

Radiation pressure,

$P=\dfrac{I}{c}(1+r)$ . . . . .(1)

Intensity,

$I=\dfrac{p}{4\pi r^2}=\dfrac{6\times 10^5}{4\times 3.14 \times 2.5\times 2.5}$

$I=7.64\times 10^4W/m^2$

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

$r=50$ %

$=0.5$ reflectivity

From equation (1)

$P=\dfrac{7.64\times 10^4}{3\times 10^8}(1+0.5)$

$P=4 \times 10^{-5} Pa$

A hydrogen atom, originally in its ground state, absorbs a photon and goes into an excited state.
The atom will then most likely

0%
be ionized
0%
emit a photon
0%
emit an electron
0%
always be in that excited state
0%
undergo nuclear fission

What should be the De-Broglie wavelength of confined particle in the box ?

0%
L/2n
0%
2L/n
0%
L/n
0%
nL

$\sqrt{V}$ on two particles

$A$ and

$B$ are plotted against de-Broglie wavelength. Where

$V$ is the potential on the particle. Which

0%
$M_A=m_B$
0%
$M_A > m_B$
0%
$M_A < m_B$
0%
$M_A \leq m_B$

The anode plate in an experiment on photoelectric effect is kept vertically above the cathode plate. Light source is put on and a saturation photocurrent is recorded. An electric field is switched on which has vertically downward direction.

0%
no change in the photocurrent
0%
The kinetic energy of the electrons will increase
0%
The stopping potential will decrease
0%
The threshold wavelength will increase

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let

$\lambda _ { n } , \lambda _ { g }$ be the de Broglie wavelength of the electron in the

$n ^ { t h }$ state and the ground state respectively. Let

$\Lambda _ { n }$ be the wavelength of the emitted photon in the transition from the

$n ^ { t h }$ state to the ground state. For large n, (A, B are constants)

0%
$\Lambda _ { n } \approx A + \frac { B } { \lambda _ { n } ^ { 2 } }$
0%
$\Lambda _ { n } \approx A + B \lambda _ { n }$
0%
$\Lambda _ { n } ^ { 2 } \approx A + B \lambda _ { n } ^ { 2 }$
0%
$\Lambda _ { n } ^ { 2 } \approx \lambda$

The momentum of a photon of energy

$1\ MeV$ in

$kg\ m/s$ , will be

0%
$0.33 \times 10 ^ { 6 }$
0%
$7 \times 10 ^ { - 24 }$
0%
$10 ^ { - 22 }$
0%
$5 \times 10 ^ { - 22 }$

The de Broglie wavelength of an electron moving with a velocity

$1.5\times 10^8ms^{-1}$ is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of photon is?(apply non relativistic formula for electron)

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

Explanation

${ K }_{ particle }=\frac { 1 }{ 2 } m{ v }^{ 2 }\quad \lambda =\frac { h }{ mv }$

${ K }_{ particle }=\frac { 1 }{ 2 } \left( \frac { h }{ \lambda v } \right) .{ v }^{ 2 }=\frac { vh }{ 2\lambda } ........(1)$

${ K }_{ photon }=\frac { hc }{ \lambda } ........(2)$

$\frac { { K }_{ particle } }{ { K }_{ photon } } =\frac { v }{ 2c } =\frac { 1.5\times { 10 }^{ 8 } }{ 2\times 3\times { 10 }^{ 8 } } =\frac { 1 }{ 4 }$

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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