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

A negatively charged electroscope with zinc disc discharges when irradiated by an ultraviolet lamp. What caused this?
  • $$\alpha -$$ particles from the source combine with electrons of the disc
  • electrons escape from the disc when ultraviolet radiation falls on it
  • ultraviolet rays ionize the air surrounding the electroscope
  • 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:
  • neutron
  • proton
  • electron
  • deuteron
The correct curve between the energy of photon (E) and its wavelength ( $$ \lambda $$ ) is
With the decrease in the wave length of the incident radiation the velocity of the photoelectrons emitted from a given metal
  • Remains same
  • Increases
  • Decreases
  • Increases first and then decreases
The mass of a photon in motion is (given its frequency $$=$$ x ):
  • $$\dfrac{hx}{c^{2}}$$
  • $$hx^{3}$$
  • $$\dfrac{hx^{3}}{c^{2}}$$
  • Zero
The deBroglie wavelength associated with a particle of mass m, moving with a velocity $$'v\ '$$ and energy E is given by
  • $$\dfrac{h}{mv^{2}}$$
  • $$\dfrac{mv}{h^{2}}$$
  • $$\dfrac{h}{\sqrt{2mE}}$$
  • $$\sqrt{2mE}/h$$
An electron of charge 'e' and mass 'm' is accelerated from rest by a potential difference 'V'. The de Broglie wavelength is
  • Directly proportional to the square root of potential difference.
  • Inversely proportional to the square root of potential difference
  • Directly proportional to the square root of electron mass
  • Inversely proportional to the cube root of electron mass
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
  • $$\dfrac{12.27}{\sqrt{V}}A^{0}$$
  • $$\dfrac{12.27}{V}A^{0}$$
  • $$12.27\sqrt{V}A^{0}$$
  • $$\dfrac{1}{12.27\sqrt{V}}A^{0}$$
The de Broglie wavelength of a molecule of thermal energy KT (K is Boltzmann constant and T is absolute temperature) is given by :
  • $$\dfrac{h}{\sqrt{2mKT}}$$
  • $$\dfrac{h}{2mKT}$$
  • $$h\sqrt{2mKT}$$
  • $$\dfrac{1}{h\sqrt{2mKT}}$$
The graph between the de Broglie wavelength and the momentum of a photon is a 
  • Rectangular hyperbola
  • Circle
  • Parabola
  • Straight line
The wavelength associated with a photon of energy 3.31 eV is nearly
  • 4000 $$A^{o}$$
  • 3750 $$A^{o}$$
  • 5000 $$A^{o}$$
  • 400 $$A^{o}$$
Let $$p$$ and $$E$$ denote the linear momentum and energy, respectively, of a proton. If the wavelength is decreased
  • Both $$p$$ and $$E$$ increase
  • $$p$$ increases and $$E$$ decreases
  • $$p$$ decreases and $$E$$ increases
  • Both $$p$$ and $$E$$ decrease
The ratio of the wavelengths of a photon and that of an electron of same energy $$E$$ will be [$$m$$ is mass of electron]:
  • $$\sqrt{\dfrac{2m}{E}}$$
  • $$\sqrt{\dfrac{E}{2m}}$$
  • $$c\sqrt{\dfrac{2m}{E}}$$
  • $$\sqrt{\dfrac{Ec}{2m}}$$
The wavelengths of a proton and a photon are same. Then :
  • their velocities are same
  • their momenta are equal
  • their energies are same
  • their speeds are same
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
  • $$\lambda_{e}=\lambda _{p}$$
  • $$\lambda_{e}> \lambda _{p}$$
  • $$\lambda_{e}< \lambda _{p}$$
  • $$\lambda_{e} \leq \lambda _{p}$$
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 :
  • 0.9 eV
  • 2. 2 eV
  • 4. 4 eV
  • 3.1 eV
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
  • $$\lambda_{e} = \lambda _{p}$$
  • $$\lambda_{e} < \lambda _{p}$$
  • $$\lambda_{e} > \lambda _{p}$$
  • $$\lambda_{e} \geq \lambda _{p}$$
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 :
  • $$7.5\times 10^{15}eV,2.7eV$$
  • $$6.5\times 10^{16}eV,2.9eV$$
  • $$6.5\times 10^{16}eV,2.7eV$$
  • $$7.5\times 10^{16}eV,1.9eV$$
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
  • 1 : 3
  • 3 : 1
  • 1 : 2
  • 1 : 4
The energy that should be added to an electron to reduce its de-Broglie wavelength from 1nm to 0.5nm is
  • Four times the intial energy
  • Equal to initial energy
  • Twice the initial energy
  • Thrice the intial energy
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:
  • $$\lambda_{e},\lambda _{p},\lambda_{n} ,\lambda_{\alpha } $$
  • $$\lambda_{\alpha },\lambda _{n},\lambda_{p} ,\lambda_{e}$$
  • $$\lambda_{e},\lambda _{n},\lambda_{p} ,\lambda_{\alpha }$$
  • $$\lambda_{p},\lambda _{e},\lambda_{\alpha } ,\lambda_{n }$$
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)
  • 140 $$A^{0}$$
  • 0.14 $$A^{0}$$
  • 14 $$A^{0}$$
  • 1.4$$A^{0}$$
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:
  • $$6.6\times 10^{-8}cm$$
  • $$0.66\times 10^{-8}cm$$
  • $$6.6\times 10^{-8}m$$
  • $$10cm$$
Two particles of masses m and 2m have equal kinetic energies. Their de Broglie wavelengths are in the ratio of 
  • $$1 :1$$
  • $$1:2$$
  • $$1:\sqrt{2}$$
  • $$\sqrt{2}:1$$
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 :
  • $$\dfrac{1}{8}$$
  • $$\dfrac{3}{8}$$
  • $$\dfrac{5}{8}$$
  • $$\dfrac{7}{8}$$
If electron is having a wavelength of 100 $$A^{0}$$, then momentum is $$(gm \  cm \ s^{-1})$$ units
  • $$6.6\times 10^{-32}$$
  • $$6.6\times 10^{-29}$$
  • $$6.6\times 10^{-35}$$
  • $$6.6\times 10^{-21}$$
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)
  • 4.8 $$A^{0}$$
  • 3.6 $$A^{0}$$
  • 1.2 $$A^{0}$$
  • 2.4 $$A^{0}$$
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 :
  • $$1:2\sqrt{2}$$
  • $$2:1$$
  • $$2\sqrt{2}:1$$
  • $$4:1$$
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 :
  • 1 : 1837
  • 43 : 1
  • 1837 : 1
  • 1 : 43
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)
  • $$6.6\times 10^{-26}kg$$ m/s
  • $$6.6\times 10^{-25}kg$$ m/s
  • $$6.6\times 10^{-24}kg$$ m/s
  • $$6.6\times 10^{-22}kg$$ m/s
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.$$\sqrt{\dfrac{0.0409}{V}}$$
.
  • a-g, b-e, c-h, d-f
  • a-g, b-h, c-f, d-e
  • a-h, d-e, c-f, d-g
  • a-h, h-f, c-e, d-h
If $$\lambda_{0} $$ 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
  • $$2\sqrt{2}\lambda _{0}$$
  • $$\dfrac{\lambda _{0}}{2}$$
  • $$\dfrac{\lambda _{0}}{2\sqrt{2}}$$
  • $$\dfrac{\lambda _{0}}{\sqrt{2}}$$
If the energy of a particle is reduced to one fourth, then the percentage increase in its de Broglie wavelength will be :
  • $$4$$1%
  • $$141$$%
  • $$100$$%
  • $$71$$%
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_________
  • $$1/12$$
  • $$6/7$$
  • $$5/12$$
  • $$5/7$$
If the velocity of a particle is increased three times, then the percentage decrease in its de Broglie wavelength will be :
  • $$33.3$$%
  • $$66.6$$%
  • $$99.9$$%
  • $$22.2$$%
A monochromatic source of light operating at 200 W emits $$4 \times 10^{20}$$ photons/second. Then the wavelength of light used is
  • 3000 $$A^{0}$$
  • 5000 $$A^{0}$$
  • 4000 $$A^{0}$$
  • 6000$$A^{0}$$
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
  • $$\lambda$$
  • $$2\lambda$$
  • $$\lambda/2$$
  • $$\lambda/4$$
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
  • $$p_{m}/200$$
  • $$p_{m}/100$$
  • $$201p_{m}$$
  • $$100p_{m}$$
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)
  • $$2.5\times 10^{11}$$
  • $$5.0\times 10^{11}$$
  • $$3.5\times 10^{11}$$
  • $$5.0\times 10^{10}$$
Work function of a metal is 2.1 eV. The pair of wavelengths which is able to emit photo-electrons is
  • 4000 $$A^{0}$$, 7500 $$A^{0}$$
  • 5500 $$A^{0}$$, 6000 $$A^{0}$$
  • 4000 $$A^{0}$$, 5000 $$A^{0}$$
  • None of these
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]
  • $$\dfrac{M}{m}$$
  • $$\sqrt{\dfrac{M}{m}}$$
  • $$\dfrac{m}{M}$$
  • $$\sqrt{\dfrac{m}{M}}$$
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 
  • 1.5$$A^{0}$$
  • 1.0$$A^{0}$$
  • 3.0 $$A^{0}$$
  • 0.5 $$A^{0}$$
The de Broglie wavelength associated with an electron of energy 500 eV is given by
(take $$h=6.63\times 10^{-34}Js,m=9.11\times 10^{-31}kg$$ )
  • 0.28$$A^{0}$$
  • 1.410 $$A^{0}$$
  • 0.66 $$A^{0}$$
  • 0.55 $$A^{0}$$
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 $$(h = 6.6 \times 10^{-34} Js, c = 3 \times 10^{8} m/s)$$ 
  • $$3.0\times 10^{3}Hz$$
  • $$6.0\times 10^{3}Hz$$
  • $$7.5\times 10^{12}Hz$$
  • $$1.5\times 10^{13}Hz$$
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.)
  • $$2.42 \times10^{-12}$$m
  • $$4.24 \times10^{-12}$$m
  • $$1.82 \times10^{-12}$$m
  • $$1.28 \times10^{-12}$$m
The momentum ( in Kg-m/s) of an electron having wavelength 2$$A^{0}$$  $$(h=6.62 \times10^{-34}Js.)$$
  • $$3.3125\times 10^{-24}$$
  • $$4.24 \times 10^{-23}$$
  • $$1.82 \times 10^{-29}$$
  • $$1.28 \times 10^{-12}$$
The momentum of a photon having energy equal to the rest energy of an electron is
  • Zero
  • $$2.7\times 10^{-22}kgms^{-1}$$
  • $$1.99\times 10^{-24}kgms^{-1}$$
  • $$\infty$$
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
  • 1
  • $$\dfrac{m_{e}}{m_{p}}$$
  • $$\dfrac{m_{p}}{m_{e}}$$
  • $$\sqrt{\dfrac{m_{p}}{m_{e}}}$$
The de-Broglie wavelength associated with an electron accelearated to potential difference of V volt is:
  • $$V\times 10^{-10}m$$
  • $$\sqrt{\dfrac{150}{V}}A^{0}$$
  • $$\dfrac{150}{V}A^{0}$$
  • $$150V\times 10^{-8}m$$
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
  • $$\dfrac{V}{8}$$ volt
  • 4V volt
  • 2V volt
  • 1838V volt
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