MCQ Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 3

A negatively charged electroscope with zinc disc discharges when irradiated by an ultraviolet lamp. What caused this?

0%
$\alpha -$ particles from the source combine with electrons of the disc
0%
electrons escape from the disc when ultraviolet radiation falls on it
0%
ultraviolet rays ionize the air surrounding the electroscope
0%
the disc becomes hot and thermionic emission takes place

Which of the following particles - neutron, proton,electron and deuteron has the lowest energy if all have the same de Broglie wavelength:

0%
neutron
0%
proton
0%
electron
0%
deuteron

Explanation

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

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

So

$E\alpha \dfrac{1}{m}$

Deuteron is mass of 1 proton and 1 neutron ,hence its mass is maximum out of the given particles ,so its energy is the lowest.

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

$\lambda$ ) is

Explanation

Energy of photon

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

$\Longrightarrow E \propto \dfrac { 1 }{ \lambda }$
So, it will be a rectangular hyperbola.

With the decrease in the wave length of the incident radiation the velocity of the photoelectrons emitted from a given metal

0%
Remains same
0%
Increases
0%
Decreases
0%
Increases first and then decreases

The mass of a photon in motion is (given its frequency

$=$ x ):

0%
$\dfrac{hx}{c^{2}}$
0%
$hx^{3}$
0%
$\dfrac{hx^{3}}{c^{2}}$
0%
Zero

Explanation

By Einstein's mass energy equivalence

$E=mc^{2}$
but

$E=hx$

$\therefore hx=mc^{2}$

$\Rightarrow m=\dfrac{hx}{c^{2}}$

The deBroglie wavelength associated with a particle of mass m, moving with a velocity

$'v\ '$ and energy E is given by

0%
$\dfrac{h}{mv^{2}}$
0%
$\dfrac{mv}{h^{2}}$
0%
$\dfrac{h}{\sqrt{2mE}}$
0%
$\sqrt{2mE}/h$

Explanation

We know,

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

$mv=\sqrt{2Em}$
so debroglie wavelength

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

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

An electron of charge 'e' and mass 'm' is accelerated from rest by a potential difference 'V'. The de Broglie wavelength is

0%
Directly proportional to the square root of potential difference.
0%
Inversely proportional to the square root of potential difference
0%
Directly proportional to the square root of electron mass
0%
Inversely proportional to the cube root of electron mass

Explanation

$\frac{1}{2}mv^{2}=eV$

$mv=\sqrt{2eVm}$
So de-broglie wavelength

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

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

$\therefore \lambda \alpha \frac{1}{\sqrt{V}}$

An electron of mass

$9.1\times 10^{-31}$ kg and charge

$1.6\times 10^{-19}$ C is accelerated through a potential difference of 'V' volt. The de Broglie wavelength

$(\lambda )$ associated with the electron is

0%
$\dfrac{12.27}{\sqrt{V}}A^{0}$
0%
$\dfrac{12.27}{V}A^{0}$
0%
$12.27\sqrt{V}A^{0}$
0%
$\dfrac{1}{12.27\sqrt{V}}A^{0}$

Explanation

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

$mv=\sqrt{2meV}$
Now

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

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

$=\dfrac{12.27}{\sqrt{V}}A^{0}$

The de Broglie wavelength of a molecule of thermal energy KT (K is Boltzmann constant and T is absolute temperature) is given by :

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

Explanation

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

$mv=\sqrt{2mKT}$

So de-broglie wavelength

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

$\rightarrow\lambda=\dfrac{h}{\sqrt{2mKT}}$

The graph between the de Broglie wavelength and the momentum of a photon is a

0%
Rectangular hyperbola
0%
Circle
0%
Parabola
0%
Straight line

Explanation

The relationship between de Broglie wavelength

$(\lambda)$ and the momentum (p) of a photon is given by:

$\lambda=\dfrac{h}{p}$ .
So, the graph between the de Broglie wavelength and the momentum of a photon is a rectangular hyperbola.

The wavelength associated with a photon of energy 3.31 eV is nearly

0%
4000 $A^{o}$
0%
3750 $A^{o}$
0%
5000 $A^{o}$
0%
400 $A^{o}$

Explanation

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

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

$=\dfrac{1240}{3.31} \ \ \ (hc=1240 eV)$

$=375 nm$

$=3750 A^o$
So, the answer is option (B).

Let

$p$ and

$E$ denote the linear momentum and energy, respectively, of a proton. If the wavelength is decreased

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

Energy

$E=h\nu=\dfrac{hc}{\lambda}$ and momentum

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

Thus energy and momentum are inversely proportional to wavelength.

So when wavelength

$(\lambda)$ is decreased, both energy E and momentum p will increase.

The ratio of the wavelengths of a photon and that of an electron of same energy

$E$ will be [

$m$ is mass of electron]:

0%
$\sqrt{\dfrac{2m}{E}}$
0%
$\sqrt{\dfrac{E}{2m}}$
0%
$c\sqrt{\dfrac{2m}{E}}$
0%
$\sqrt{\dfrac{Ec}{2m}}$

Explanation

Wavelength of photon:

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

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

Wavelength of electron:

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

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

$mv=\sqrt{2Em}$

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

$\dfrac{\lambda _{p}}{\lambda _{e}}=c\sqrt{\dfrac{2m}{E}}$

The wavelengths of a proton and a photon are same. Then :

0%
their velocities are same
0%
their momenta are equal
0%
their energies are same
0%
their speeds are same

Explanation

For proton,

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

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

For photon,

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

$\lambda _{2}=\dfrac{h}{E/c}$

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

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

An electron and a proton possess the same amount of Kinetic energy. Then the relation between the wavelength of electron

$(\lambda _{e})$ and the wavelength of proton

$(\lambda _{p})$ is

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

Explanation

For an electron

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

$mv=\sqrt{2Em}$

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

$=\dfrac{h}{\sqrt{2Em_{e}}}$

For a proton

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

$mv=\sqrt{2Em}$

$\lambda _{p}=\dfrac{h}{\sqrt{2Em_{p}}}$

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

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

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

The energy of emitted photoelectrons from a metal is 0.9 eV. The work function of the metal is 2.2 eV. Then the energy of the incident photon is :

0%
0.9 eV
0%
2. 2 eV
0%
4. 4 eV
0%
3.1 eV

Explanation

Energy of incident photon

$=$ energy of emitted photoelectrons

$+$ work function

$=2.9eV+2.2eV$

$=3.1eV$

So, the answer is (D).

A proton and an electron are accelerated by the same potential difference. Let

$\lambda_{e}$ and

$\lambda_{p}$ denote the de Broglie wavelength of the electron and the proton respectively, then

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

Explanation

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

$mv=\sqrt{2meV}$

So de Broglie wavelength

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

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

So

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

$\lambda _{p}=\dfrac{h}{\sqrt{2m_{p}eV}}$

So,

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

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

$652$ nm in pulses that are of

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

$0.6W$ . The energy in each pulse and in a single 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%
$7.5\times 10^{16}eV,1.9eV$

Explanation

Energy of photon

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

$=\dfrac{1240 eV-nm}{652 nm} \ \ \ \ (hc=1240 eV-nm)$

$=1.9eV$

Energy

$=$ Power

$\times$ time

$=0.6\times 20\times 10^{-3}\times \dfrac{1}{1.6\times 10^{-16}}$

$=7.5\times 10^{16}eV$

Two photons have energies of

$4.95\times 10^{-19}$ J and

$14.85\times 10^{-19}$ J. Then the ratio of their wavelengths is

0%
1 : 3
0%
3 : 1
0%
1 : 2
0%
1 : 4

Explanation

energy (E)

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

So,

$\dfrac{E_1}{E_2} = \dfrac{\lambda_2}{\lambda_1}$

$\dfrac{\lambda_1}{\lambda_2} = \dfrac{E_2}{E_1}$

$= \dfrac{14.85*10^{-19}}{4.95*10^{-19}}$

$= \dfrac{3}{1}$

So, the answer is option (B).

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

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

Explanation

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

$mv=\sqrt{2Em}$

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

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

$\lambda _{1}=\dfrac{h}{\sqrt{2E_{1}m}}$

$\lambda _{2}=\dfrac{h}{\sqrt{2E_{2}m}}$

Given, $\lambda_{1}=2\lambda_{2}$

$\dfrac{h}{\sqrt{2E_{1}m}}=2\dfrac{h}{\sqrt{2E_{2}m}}$

$\dfrac{1}{E_{1}}=\dfrac{4}{E_{2}}$

$E_{2}=4E_{1}$

$E_{2}=E_{1}+3E_{1}$

The magnitude of the De-Broglie wavelength (

$\lambda$ ) of an electron (e),proton(p),neutron (n) and

$\alpha$ - particle (

$\alpha$ ) all having the same energy of MeV, in the increasing order will follow the sequence:

0%
$\lambda_{e},\lambda _{p},\lambda_{n} ,\lambda_{\alpha }$
0%
$\lambda_{\alpha },\lambda _{n},\lambda_{p} ,\lambda_{e}$
0%
$\lambda_{e},\lambda _{n},\lambda_{p} ,\lambda_{\alpha }$
0%
$\lambda_{p},\lambda _{e},\lambda_{\alpha } ,\lambda_{n }$

Explanation

We know

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

$mv=\sqrt{2Em}$

So

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

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

$\lambda _{p}=\dfrac{h}{\sqrt{2Em_{p}}}$

$\lambda _{n}=\dfrac{h}{\sqrt{2Em_{n}}}$

$\lambda _{\alpha }=\dfrac{h}{\sqrt{2Em_{\alpha }}}$

we know

$m_{\alpha }> m_{n}> m_{p}> me$
So

$\lambda _{\alpha }< \lambda _{n}< \lambda _{p}< \lambda e$

The de Broglie wavelength of an electron having 80 eV of energy is nearly

(

$1eV=1.6\times10^{-19}J$ , Mass of electron

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

Planck’s constant

$=6.6\times 10^{-34}$ Js) (nearly)

0%
140 $A^{0}$
0%
0.14 $A^{0}$
0%
14 $A^{0}$
0%
1.4 $A^{0}$

Explanation

$given,\ \dfrac{1}{2}mv^{2}= 80\times 1.6\times 10^{-19}$

$mv=\ \sqrt{2\times 9\times 10^{-3.1}\times 80\times 1.6\times 10^{-19}}$

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

$=\dfrac{6.6\times 10^{34}}{\sqrt{2\times 9\times 10^{-31}\times 80\times 1.6\times 10^{-19}}}$

$=\ 1.4\ A^{0}$

So, the answer is option (D).

A particle of mass

$10^{-31}$ kg is moving with a velocity equal to

$10^{5} ms^{-1}$ . The wavelength of the particle is equal to:

0%
$6.6\times 10^{-8}cm$
0%
$0.66\times 10^{-8}cm$
0%
$6.6\times 10^{-8}m$
0%
$10cm$

Explanation

We know that

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

$\lambda =\ \dfrac{6.63\times 10^{-34}}{10^{-31}\times 10^5}$

$\lambda =6.6\times 16^{-8}m$

So, the answer is option (C).

Two particles of masses m and 2m have equal kinetic energies. Their de Broglie wavelengths are in the ratio of

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

Explanation

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

$mv=\sqrt{2Em}$
De-broglie wavelength

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

$\lambda_m=\dfrac{h}{\sqrt{2E\times m}}$

$\lambda_{2m} =\dfrac{h}{\sqrt{2\times E\times 2m}}$

$\dfrac{\lambda_m}{\lambda_{2m}}=\dfrac{\sqrt{2}}{1}$

So, the answer is option (D).

The de-Broglie wavelength of a particle moving with a velocity

$2.25 \times 10^{8}\ m/s$ is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is :

0%
$\dfrac{1}{8}$
0%
$\dfrac{3}{8}$
0%
$\dfrac{5}{8}$
0%
$\dfrac{7}{8}$

Explanation

$K_{particle }=\dfrac 12 mv^2$ also

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

$\Rightarrow K_{particle }=\dfrac 12 \left( \dfrac {h}{\lambda v}\right). v^2=\dfrac {vh}{2\lambda}$ ...(i)

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

$\therefore \dfrac {K_{particle }}{K_{photon}}=\dfrac {v}{2c}=\dfrac {2.25\times 10^8}{2\times 3\times 10^8}=\dfrac 38$

If electron is having a wavelength of 100

$A^{0}$ , then momentum is

$(gm \ cm \ s^{-1})$ units

0%
$6.6\times 10^{-32}$
0%
$6.6\times 10^{-29}$
0%
$6.6\times 10^{-35}$
0%
$6.6\times 10^{-21}$

Explanation

Momentum =

$\dfrac{h}{\lambda }$

$=\dfrac{6.6\times 10^{-34}}{100\times 10^{-10}}$

$=6.6\times 10^{-26}\ kg\ m/s$

$=6.6\times 10^{-26}\times 1000\times 100\ gm\ cm/s$ (

$\because$ 1kg=1000gm 1m=100cm)

$=6.6\times 10^{-21}\ gm\ cm/s.$

So, the answer is option (D).

The wavelength corresponding to a beam of electrons whose kinetic energy is 100 eV is
(

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

$1eV=1.6\times10^{-19}J$ J,

$m_{e}= 9.1 \times 10^{-31}$ kg)

0%
4.8 $A^{0}$
0%
3.6 $A^{0}$
0%
1.2 $A^{0}$
0%
2.4 $A^{0}$

Explanation

$\dfrac{1}{2}mv^{2}=\ 100eV$

$mv=\sqrt{2\times 9.1\times 10^{-31}\times 100\times 1.6\times10^{-19}}$

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

$=\ 1.2\ A^{\circ}$

So, the answer is option (C).

A proton and an alpha particle are accelerated through the same potential difference. The ratio of wavelengths associated with proton and alpha particle respectively is :

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

Explanation

We know,

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

$mv = \sqrt{2meV}$

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

$\lambda _p = \dfrac{h}{\sqrt{2m_{p}q_{p}V}}$

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

$\lambda _{\alpha} = \dfrac{h}{\sqrt{2m_{\alpha} q_{\alpha}V}}$

$= \dfrac{h}{\sqrt{2\times4m\times2e\times V}}$

$\dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{2\sqrt2}{1}$

So, the answer is option (C).

If an electron and a proton have the same kinetic energy, the ratio of the de Broglie wavelengths of proton and electron would approximately be :

0%
1 : 1837
0%
43 : 1
0%
1837 : 1
0%
1 : 43

Explanation

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

$p= mv= \sqrt{2mE}$

$= \sqrt{2mqv}$

De Broglie wavelength associated with change particle

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

where

$E = \dfrac{1}{2}mv^{2} = K.E,$ P is momentum of change

So,

$\lambda _{P} = \dfrac{h}{\sqrt{2m_{p}E}}$

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

So,

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

$= \sqrt {\dfrac{m_{e}}{m_{p}}}$

$= \sqrt {\dfrac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}}}$

$= \sqrt{5.449 \times 10^{-4}}$

$= 2.33 \times 10^{-2}$

$= 0.0233$

$= \dfrac{1}{43}$

So, the answer is option (D).

A particle having a de Broglie wavelength of 1.0

$A^{0}$ is associated with a momentum of (given

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

0%
$6.6\times 10^{-26}kg$ m/s
0%
$6.6\times 10^{-25}kg$ m/s
0%
$6.6\times 10^{-24}kg$ m/s
0%
$6.6\times 10^{-22}kg$ m/s

Explanation

Momentum of a particle with given de Broglie wavalength

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

$=\dfrac{6.6\times 10^{-34}}{1\times 10^{-10}}$

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

So, the answer is option (C).

A charged particle drops through V volts. Match the de Broglie wavelength for given particles
Particle

$\lambda\ in\ A^o$

a. Electron	e. $\sqrt{\dfrac{0.0817}{V}}$
b. Deuteron	f. $\sqrt{\dfrac{0.0102}{V}}$
c. $\alpha$ particle	g. $\sqrt{\dfrac{150}{V}}$
d. Proton	h. $\sqrt{\dfrac{0.0409}{V}}$

0%
a-g, b-e, c-h, d-f
0%
a-g, b-h, c-f, d-e
0%
a-h, d-e, c-f, d-g
0%
a-h, h-f, c-e, d-h

Explanation

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

$mv=\sqrt{2qVm}$

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

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

$\lambda _{e}=\dfrac{h}{\sqrt{2m_e\times e\times V}}$

$\lambda _{e}=\sqrt{\dfrac{150}{V}}$

$\lambda _{\alpha }=\dfrac{h}{\sqrt{2\times 4\times m_p\times 2\times e\times V}}$

$\lambda _{\alpha }=\sqrt{\dfrac{0.0102}{V}}$

$\lambda _p=\sqrt{\dfrac{0.0817}{V}}$

$\lambda_d\ =\sqrt{\dfrac{0.0409}{V}}$

$\left ( m_d=2m_p\ \ q_{d}=e \right )$

So, the answer is option (B).

$\lambda_{0}$ is the de Broglie wavelength for a proton accelerated through a potential difference of 100V, the de Broglie wavelength for

$\alpha$ -particle accelerated through the same potential difference is

0%
$2\sqrt{2}\lambda _{0}$
0%
$\dfrac{\lambda _{0}}{2}$
0%
$\dfrac{\lambda _{0}}{2\sqrt{2}}$
0%
$\dfrac{\lambda _{0}}{\sqrt{2}}$

Explanation

Proton is accelerated through a potential difference 100 V

$so,\ \dfrac{1}{2}mv^{2}= e\times 100$

$mv= \sqrt{2m_{p}e\times 100}$

$\lambda _o=\dfrac{h}{\sqrt{2me\times 100}}$

$De \ broglie \ wavelength\ of\ \alpha -\ particle$

$\lambda _{\alpha } = \dfrac{h}{\sqrt{2m_{\alpha }v_{\alpha }\times 100}}$

$\lambda _{\alpha } = \dfrac{h}{\sqrt{2\times 4m\times 2e\times 100}}$

$so\ \lambda _{\alpha} = \dfrac{\lambda _0}{2\sqrt{2}}$

So, the answer is option (C).

If the energy of a particle is reduced to one fourth, then the percentage increase in its de Broglie wavelength will be :

0%
$4$ 1%
0%
$141$ %
0%
$100$ %
0%
$71$ %

Explanation

We know,

$\dfrac{1}{2^{\cdot }}mv^{2}=E$

$mv=\sqrt{2mE}$ &

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

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

$\lambda ^{'}=\dfrac{h}{\sqrt{2m\dfrac{E}{4}}}$

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

$\lambda ^{'}=2\lambda$

So, the answer is option (C).

In a Compton effect experiment, X-ray photons of wavelength 0.22

$A^{0}$ suffer a Compton shift of 0.02

$A^{0}$ . The fractional change in the energy of the incident photons is_________

0%
$1/12$
0%
$6/7$
0%
$5/12$
0%
$5/7$

Explanation

Given, E(initial energy of photon)

$=hc/0.22$ (By the formula

$E=hc/(wavelength))$

e(final energy of photon)

$=hc/0.24$

Therefore, fractional change in energy of photon

$=(E-e)/E=1/12$

If the velocity of a particle is increased three times, then the percentage decrease in its de Broglie wavelength will be :

0%
$33.3$ %
0%
$66.6$ %
0%
$99.9$ %
0%
$22.2$ %

Explanation

Wavelength,

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

$\lambda ^{'}=\dfrac{h}{3mv}$

$\dfrac{\lambda ^{'}}{\lambda }=\dfrac{1}{3}$

$\lambda ^{'}=\dfrac{\lambda }{3}$

$\Delta \lambda =\lambda -\dfrac{\lambda }{3}$

$=\dfrac{2\lambda }{3}$

$=66.6$ % decrease in

$\lambda$

So, the answer is option (B).

A monochromatic source of light operating at 200 W emits

$4 \times 10^{20}$ photons/second. Then the wavelength of light used is

0%
3000 $A^{0}$
0%
5000 $A^{0}$
0%
4000 $A^{0}$
0%
6000 $A^{0}$

Explanation

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

$\ 200=\ \dfrac{4\times 10^{20}\times 20\times 10^{-26}}{\lambda }$

$\lambda =\ 4000\ A^{0}$

So, the answer is option (C).

The de-Broglie wavelength of a bus moving with speed 'v' is

$'\lambda'$ .Some passengers left the bus at a stoppage . Now when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. The de-Broglie wavelength now is

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

Explanation

Initially

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

Now

$\dfrac{1}{2}M(2v)^{2}=2\times \dfrac{1}{2}mv^{2}$

$2Mv^{2}= mv^{2}$

$M=\dfrac{m}{2}$

$so\ \lambda _{Now}=\dfrac{h}{\dfrac{m}{2}\times 2v}$

$=\lambda$

So, the answer is option (A).

If the momentum of an electron is changed by

$p_{m}$ , then the de Broglie wavelength associated with it increased by

$0.5\%$ . The initial momentum of electron will be

0%
$p_{m}/200$
0%
$p_{m}/100$
0%
$201p_{m}$
0%
$100p_{m}$

Explanation

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

$\lambda_i=\dfrac{h}{p_{i}}$

Now

$\lambda _{i}+\dfrac{0.5}{100}\lambda _{i}=\dfrac{h}{p-p_{m}}$

$\dfrac{201}{200}(\dfrac{h}{p})=\dfrac{h}{p-p_{m}}$

$201p-201p_{m}=200p$

$p=201p_{m}$
So, the answer is option (C).

Light of wavelength

$5000 A^o$ falls on a sensitive surface. If the surface has received

$10^{-7}J$ of energy, then the number of photons incident on the surface is nearly (given

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

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

0%
$2.5\times 10^{11}$
0%
$5.0\times 10^{11}$
0%
$3.5\times 10^{11}$
0%
$5.0\times 10^{10}$

Explanation

Energy= nhcλNumber of photons

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

where

$E = 10^{-7} \ J$ and

$\lambda = 5000\times 10^{-10} \ m$

$\therefore$

$N = \dfrac{10^{-7}\times 5000\times 10^{-10}}{(6.6\times 10^{-34})(3\times 10^8)} =2.5\times 10^{11}$

So, the answer is option (A).

Work function of a metal is 2.1 eV. The pair of wavelengths which is able to emit photo-electrons is

0%
4000 $A^{0}$ , 7500 $A^{0}$
0%
5500 $A^{0}$ , 6000 $A^{0}$
0%
4000 $A^{0}$ , 5000 $A^{0}$
0%
None of these

Explanation

The maximum wavelength required for metals to eject photons ,

$\dfrac{hC} {\lambda} = 2.1 eV$

$\Rightarrow \dfrac{4.41 \times 10^{-15} \times 3 \times 10^{8}}{\lambda} = 2.1$

$\Rightarrow \lambda = 6.3 \times 10^{-7}m \Rightarrow \lambda = 6300 A^{\circ}$
So,
Both The wavelengths should be less that

$6300 A^{\circ}$ .
So, the answer is option (C).

A positron and a proton are accelerated by the same accelerating potential. Then the ratio of the associated wavelengths of the positron and the proton will be :
[

$M$ = Mass of proton,

$m$ = Mass of positron]

0%
$\dfrac{M}{m}$
0%
$\sqrt{\dfrac{M}{m}}$
0%
$\dfrac{m}{M}$
0%
$\sqrt{\dfrac{m}{M}}$

Electrons are accelerated through a p.d. of 150V. Given

$m=9.1\times{10}^{-31}$ Js, the de Broglie wavelength associated with it is

0%
1.5 $A^{0}$
0%
1.0 $A^{0}$
0%
3.0 $A^{0}$
0%
0.5 $A^{0}$

Explanation

De-Brogtie wavelength

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

And

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

$mv=\sqrt{2meV}$

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

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

$\lambda =\dfrac{6.62\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 150}}$

$\lambda =1.0\ A^{0}$

So, the answer is option (B).

The de Broglie wavelength associated with an electron of energy 500 eV is given by

(take

$h=6.63\times 10^{-34}Js,m=9.11\times 10^{-31}kg$ )

0%
0.28 $A^{0}$
0%
1.410 $A^{0}$
0%
0.66 $A^{0}$
0%
0.55 $A^{0}$

Explanation

$\lambda =\sqrt{\dfrac{150}{V}}\ A^{\circ }$

$=\sqrt{\dfrac{150}{500}}\ A^{\circ }$

$=0.55\ A^{\circ }$

So, the answer is option (D).

The momentum of a photon of a electromagnetic radiation is

$3.3\times 10^{-29}kg \ m \ s^{-1}$ .The frequency of the associated waves is

$(h = 6.6 \times 10^{-34} Js, c = 3 \times 10^{8} m/s)$

0%
$3.0\times 10^{3}Hz$
0%
$6.0\times 10^{3}Hz$
0%
$7.5\times 10^{12}Hz$
0%
$1.5\times 10^{13}Hz$

Explanation

$p=3.3\times 10^{-29}\ kgms^{-1}$

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

$\dfrac{c}{v}=\dfrac{h}{p}$

$v=\dfrac{pc}{h}$

$=\dfrac{3.3\times 10^{-29}\times 3\times 10^{8}}{6.6\times 10^{-34}}$

$=1.5\times 10^{13}Hz$

So, the answer is option (D).

The wavelength of de broglie waves associated with a beam of protons of kinetic energy

$5 \times 10^{2}$ eV.
(Mass of each photon

$= 1.67 \times 10^{-27}$ Kg,

$h=6.62 \times 10^{-34}$ Js.)

0%
$2.42 \times10^{-12}$ m
0%
$4.24 \times10^{-12}$ m
0%
$1.82 \times10^{-12}$ m
0%
$1.28 \times10^{-12}$ m

Explanation

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

$mv=\sqrt{2Em}$

$=\sqrt{2\times 5\times 10^{2}\times 1.6\times 10^{-19}\times 1.6\times 10^{-27}}$

$=51.69\times 10^{23}\ N$

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

$=\dfrac{6.62\times 10^{-34}}{51.63\times 10^{-23}}$

$=1.28\times 10^{-12}\ m$

So, the answer is option (D).

The momentum ( in Kg-m/s) of an electron having wavelength 2

$A^{0}$

$(h=6.62 \times10^{-34}Js.)$

0%
$3.3125\times 10^{-24}$
0%
$4.24 \times 10^{-23}$
0%
$1.82 \times 10^{-29}$
0%
$1.28 \times 10^{-12}$

Explanation

The relationship between de Broglie wavelength

$(\lambda)$ and the momentum (p) of a photon is given by:

$\lambda=\dfrac{h}{p}$ .
Using this, the momentum of an electron having wavelength

$2 \mathring A$ is

$3.3125 \times 10^{-24}$ kg-m/s.
So, the answer is option (A).

The momentum of a photon having energy equal to the rest energy of an electron is

0%
Zero
0%
$2.7\times 10^{-22}kgms^{-1}$
0%
$1.99\times 10^{-24}kgms^{-1}$
0%
$\infty$

Explanation

Momentum of a particle having rest mass energy zero like photon is given by:

$E=pc$ , where p is the momentum and c is the speed of the photon.
So,

$p=E/c=\dfrac{m_0 c^2}{c}=m_0 c=2.7 \times 10^{-22}$

So, the answer is option (B).

An electron and a proton are accelerated through the same potential difference. The ratio of their de Broglie wavelengths (

$\dfrac{\lambda_{e}}{\lambda_{p}}$ ) is

0%
1
0%
$\dfrac{m_{e}}{m_{p}}$
0%
$\dfrac{m_{p}}{m_{e}}$
0%
$\sqrt{\dfrac{m_{p}}{m_{e}}}$

Explanation

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

$\lambda _{p}=\dfrac{h}{\sqrt{2m_{p}eV}}$

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

So, the answer is option (D).

The de-Broglie wavelength associated with an electron accelearated to potential difference of V volt is:

0%
$V\times 10^{-10}m$
0%
$\sqrt{\dfrac{150}{V}}A^{0}$
0%
$\dfrac{150}{V}A^{0}$
0%
$150V\times 10^{-8}m$

Explanation

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

$p=\sqrt{2meV}$

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

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

$=\sqrt{\dfrac{150}{V}} A^o$

So, the answer is option (B).

A proton when accelerated through a potential difference of V volt has wavelength

$\lambda$ associated with it .An electron to have the same

$\lambda$ must be accelerated through a p.d of

0%
$\dfrac{V}{8}$ volt
0%
4V volt
0%
2V volt
0%
1838V volt

Explanation

The energy gained by the proton will be

$E = q_oV = \dfrac{p^2}{2m_p}$ ...... ( 1 )

where

$q_o$ is the charge on the proton, V is the p.d applied, p is the momentum of the proton and

$m_p$ is the mass of the proton.

The momentum p of the proton is given by de Broglie's equation as

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

Therefore, for an electron to have the same

$\lambda$ , it must have the same momentum.

$\therefore$

$E_e = \dfrac{p^2}{2m_e}$

Using p from equation 1,

$\implies$

$E_e = \dfrac{m_pq_oV}{m_e}$

$\implies$

$E_e = q_oV(\dfrac{m_p}{m_e}) = q_o(1838V)$

Therefore, an electron must be accelerated through a p.d of 1838V volt in order to achieve a wavelength

$\lambda$ .

CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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