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JEE • PHYSICS • DUAL NATURE OF RADIATION AND MATTER

Dual nature of radiation and matter practice · Medium

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Showing 181-210 of 303 questions

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181 Medium
181 of 303
Understanding +4 | -1
Assertion: Photoelectric effect supports the quantum nature of light.

Reason: Photoelectric emission is instantaneous.
View answer and explanation
Correct option: B
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion
Option B is correct because both the assertion and reason are true, but the reason doesn't fully explain why the photoelectric effect supports the quantum nature of light.
182 Medium
182 of 303
Application +4 | -1
When a metallic surface is illuminated with monochromatic light of wavelength λ\lambda, stopping potential for photoelectric current is 3 V03 \mathrm{~V}_0. When the same metallic surface is illuminated with a light of wavelength 2λ2 \lambda, the stopping potential is V0V_0. The threshold wavelength for the surface is-
View answer and explanation
Correct option: B
4λ4 \lambda
The correct answer is B (4λ4 \lambda) because the threshold wavelength is calculated using the photoelectric effect equations and the given stopping potentials.
183 Medium
183 of 303
Application +4 | -1
An electron is accelerated through a potential difference of VV volt. The speed of electrons will be
View answer and explanation
Correct option: B
2eVm\sqrt {\frac{{2eV}}{m}}
Option B (2eVm\sqrt{\frac{{2eV}}{m}}) is correct because it is derived from equating the kinetic energy gained by the electron to the work done by the potential difference.
184 Medium
184 of 303
Application +4 | -1
Two sources of light emit with a power of 200 W200 {~W}. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm{nm} and 500 nm500 {~nm} respectively, will be :
View answer and explanation
Correct option: B
3:53: 5
Option B (3:53: 5) is correct because it is the ratio of the number of photons calculated using the power and wavelength relationship.
185 Medium
185 of 303
Understanding +4 | -1
Which of the following statement about photon is incorrect?
View answer and explanation
Correct option: A
Photons exert no pressure
Option A is correct because photons exert pressure due to their momentum, as described by the relationship between energy, momentum, and the speed of light.
186 Medium
186 of 303
Application +4 | -1
Ultraviolet light of wavelength 280 nm280 \mathrm{~nm} is used in an experiment on photoelectric effect with lithium (φ=2.5 eV)(\varphi=2.5 \mathrm{~eV}) cathode. Find

(a) the maximum kinetic energy of the photoelectrons and

(b) the stopping potential.
View answer and explanation
Correct option: A
1.9 eV,1.9 V1.9 \mathrm{~eV}, 1.9 \mathrm{~V}
Option A (1.9 eV,1.9 V1.9 \mathrm{~eV}, 1.9 \mathrm{~V}) is correct because the maximum kinetic energy is the difference between the photon energy and the work function, and the stopping potential is equal to the maximum kinetic energy in electron volts.
187 Medium
187 of 303
Remembering +4 | -1
Which of one is correct
View answer and explanation
Correct option: A
E2=p2c2{E^2} = {p^2}{c^2}
Option A (E2=p2c2{E^2} = {p^2}{c^2}) is correct because it represents the relativistic energy-momentum relation for a massless particle or the general relation when rest mass is zero.
188 Medium
188 of 303
Analyzing +4 | -1
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
View answer and explanation
Correct option: C
More than that of a proton
Option C (More than that of a proton) is correct because for the same de-Broglie wavelength, the kinetic energy is inversely proportional to the mass, and the electron has a smaller mass than the proton.
189 Medium
189 of 303
Application +4 | -1
The wavelength of light in the visible region is about 390 nm390 \mathrm{~nm} for violet colour. What are the energies of photons in (eV)(\mathrm{eV}) at the violet end. [Take h=6.63×10−34 Jsh=6.63 \times 10^{-34} \mathrm{~Js} and 1 eV=1.6×10−19J]1 \mathrm{~eV}=1.6 \times 10^{-19} \mathrm{J}].
View answer and explanation
Correct option: A
3.19 eV3.19 \mathrm{~eV}
Option A (3.19 eV3.19 \mathrm{~eV}) is correct because it is the energy of a photon with a wavelength of 390 nm, calculated using E=hcλE = \frac{{hc}}{\lambda}.
190 Medium
190 of 303
Application +4 | -1
The ratio of the de-Broglie wavelength of an α\alpha-particle and a proton of same kinetic energy is
View answer and explanation
Correct option: A
1:21:2
Option A (1:21:2) is correct because the de-Broglie wavelength is inversely proportional to the square root of the mass, and the alpha particle's mass is four times that of the proton.
191 Medium
191 of 303
Understanding +4 | -1
In which of the following cases the heavier of the two particle has a same de Broglie wavelength? The two particles :-
View answer and explanation
Correct option: B
Move with the same linear momentum
Option B (Move with the same linear momentum) is correct because de Broglie wavelength is inversely proportional to momentum, so equal momentum implies equal wavelength.
192 Medium
192 of 303
Understanding +4 | -1
When ultraviolet radiation is incident on a surface, no photoelectrons are emitted. If a second beam causes photoelectrons to be ejected, it may consist of
View answer and explanation
Correct option: B
X-rays
Option B is correct because X-rays have higher energy (and frequency) than ultraviolet radiation, so they can cause photoelectrons to be ejected when ultraviolet radiation cannot.
193 Medium
193 of 303
Application +4 | -1
The ratio of the energy of a photon with λ=150\lambda = 150 nm to that with λ=300\lambda = 300 nm is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the energy of a photon is inversely proportional to its wavelength, so halving the wavelength doubles the energy.
194 Medium
194 of 303
Understanding +4 | -1
Photoelectric effect can be explained by
View answer and explanation
Correct option: D
Quantum theory of light
Option D (Quantum theory of light) is correct because the photoelectric effect is explained by the quantum theory, where light is considered as packets of energy called photons.
195 Medium
195 of 303
Remembering +4 | -1
The energy of an electron in the nth {n^{th}}\, Bohr’s orbit is proportional to
View answer and explanation
Correct option: C
n−2{n^{ - 2}}
Option C (n−2{n^{ - 2}}) is correct because the energy of an electron in the nth Bohr orbit is inversely proportional to the square of the principal quantum number n.
196 Medium
196 of 303
Application +4 | -1
If the uncertainty in the position of an electron is 10−10{10^{ - 10}}m, then the value of uncertainty in its momentum (in kg-ms−1\mathrm{m}{\mathrm{s}^{ - 1}}) will be
View answer and explanation
Correct option: B
1.03×10−241.03 \times {10^{ - 24}}
Option B (1.03×10−241.03 \times {10^{ - 24}}) is correct because it is calculated using the Heisenberg uncertainty principle.
197 Medium
197 of 303
Analyzing +4 | -1
A ultra violet photon and an electron have the same de Broglie wavelengths, then the energy of the photon is -
View answer and explanation
Correct option: A
greater than the kinetic energy of the electron
Option A (greater than the kinetic energy of the electron) is correct because the energy of a photon with the same de Broglie wavelength is greater than the kinetic energy of the electron.
198 Medium
198 of 303
Understanding +4 | -1
If gE{\mathrm{g}_E} and gM{\mathrm{g}_M} are the acceleration due to gravity on the surfaces of the earth and the moon respectively and it Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio electronic  charge  on  the  moon  electronic  charge  on  the  earth  \frac{{\mathrm{electronic\;charge\;on\;the\;moon\;}}}{{\mathrm{electronic\;charge\;on\;the\;earth\;}}} to be
View answer and explanation
Correct option: A
1
Option A is correct because the electronic charge is a fundamental constant and is independent of the gravitational acceleration.
199 Medium
199 of 303
Application +4 | -1
A small piece of cesium metal (φ=1.9 eV)(\varphi=1.9 \mathrm{~eV}) is kept at a distance of 20 cm20 \mathrm{~cm} from a large metal plate having a charge density of 1.0×10−9C m−21.0 \times 10^{-9} \mathrm{C} \mathrm{~m}^{-2} on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm400 \mathrm{~nm} is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.
View answer and explanation
Correct option: A
22.6 eV,23.8 eV22 .6 \mathrm{~eV}, 23 .8 \mathrm{~eV}
Option A (22.6 eV,23.8 eV22.6 \mathrm{~eV}, 23.8 \mathrm{~eV}) is correct because it calculates the minimum and maximum kinetic energies of photoelectrons reaching the metal plate by considering the potential difference and the work function.
200 Medium
200 of 303
Application +4 | -1
An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is
View answer and explanation
Correct option: D
Greater than the kinetic energy of proton
Option D (Greater than the kinetic energy of proton) is correct because, for the same de-Broglie wavelength, the lighter particle (electron) has a higher kinetic energy.
201 Medium
201 of 303
Understanding +4 | -1
Assertion: The kinetic energy of photoelectrons depends on the frequency of incident radiation.

Reason: The kinetic energy of photoelectrons is determined by the difference between the energy of the incident photon and the work function of the metal.
View answer and explanation
Correct option: A
Both Assertion and Reason are true, and Reason is the correct explanation for Assertion
Option A is correct because the kinetic energy of photoelectrons is directly related to the frequency of incident radiation and the work function of the metal.
202 Medium
202 of 303
Application +4 | -1
If λp\lambda_{\mathrm{p}} and λα\lambda_\alpha be the wavelengths of protons and α\alpha-particles of equal kinetic energies, then
View answer and explanation
Correct option: D
λp=2λα\lambda_{\mathrm{p}}=2 \lambda_\alpha
Option D (λp=2λα\lambda_{\mathrm{p}}=2 \lambda_\alpha) is correct because the wavelength is inversely proportional to the square root of the mass for particles with equal kinetic energies.
203 Medium
203 of 303
Application +4 | -1
What is the de Broglie wavelength of
(a) a ball of mass 0.060 kg0.060 \mathrm{~kg} moving at a speed of 1.0 m/s1.0 \mathrm{~m} / \mathrm{s}, and
(b) a dust particle of mass 1.0×10−9 kg1.0 \times 10^{-9} \mathrm{~kg} drifting with a speed of 2.2 m/s2.2 \mathrm{~m} / \mathrm{s}?
View answer and explanation
Correct option: A
1.1×10−32 m;3.0×10−25 m1.1 \times 10^{-32} \mathrm{~m}; 3.0 \times 10^{-25} \mathrm{~m}
Option A (1.1×10−32 m;3.0×10−25 m1.1 \times 10^{-32} \mathrm{~m}; 3.0 \times 10^{-25} \mathrm{~m}) is correct because it is the de Broglie wavelength calculated for the ball and the dust particle.
204 Medium
204 of 303
Understanding +4 | -1
Electric field and magnetic field in Thomson mass spectrograph are applied
View answer and explanation
Correct option: D
Parallel simultaneously
Option D (Parallel simultaneously) is correct because in Thomson's mass spectrograph, the electric and magnetic fields are applied parallel to each other and simultaneously.
205 Medium
205 of 303
Analyzing +4 | -1
A photon and electron have same de-Broglie wavelength. Give that vv is the speed of electron and cc is the velocity of light. Ee,  Ep{E_e},\;{E_p} are the kinetic energy of electron and photon respectively. pe,  ph{p_e},\;{p_h} are the momentum of electron and photon respectively. Then which of the following relation is correct?
View answer and explanation
Correct option: A
EeEp=v2c\frac{{{E_e}}}{{{E_p}}} = \frac{v}{{2c}}
Option A is correct because the ratio of kinetic energy of electron to photon is v/2c, derived from their de-Broglie wavelengths and energy equations.
206 Medium
206 of 303
Application +4 | -1
The threshold frequency for photo electric effect on sodium corresponds to a wave length of 6000  A∘.6000\,\,{A^ \circ }. Its work function is [Joule]
View answer and explanation
Correct option: A
3.315×10−193.315 \times {10^{ - 19}}
The correct answer is 3.315×10−193.315 \times 10^{-19} because the work function is calculated using the threshold wavelength and Planck's constant.
207 Medium
207 of 303
Understanding +4 | -1
Which is not true with respect to the cathode rays
View answer and explanation
Correct option: C
Move with speed same as that of light
Option C (Move with speed same as that of light) is correct because cathode rays move at speeds significantly less than the speed of light.
208 Medium
208 of 303
Application +4 | -1
A photo cell is receiving light from a source placed at a distance of 1  m.1\;m. If the same source is to be placed at a distance of 2  m,2\;m, then the ejected electron
View answer and explanation
Correct option: D
Will be one-fourth in number
Option A (2 I2\,I) is correct because the angular momentum is conserved, and the moment of inertia doubles when the angular velocity is halved.
209 Medium
209 of 303
Analyzing +4 | -1
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector?
View answer and explanation
Correct option: C
One photon of energy 10.2 eV and an electron of energy 1.4 eV
The correct answer is C because the first photon excites the hydrogen atom, and the second photon ionizes it, releasing an electron with kinetic energy.
210 Medium
210 of 303
Application +4 | -1
A Rn photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
View answer and explanation
Correct option: D
4000  Ao4000\,\,\mathop \mathrm{A}\limits^\mathrm{o}
Option D (4000  Ao4000\,\,\mathop \mathrm{A}\limits^\mathrm{o}) is correct because the maximum wavelength that can be detected is determined by the band gap energy of the semiconductor.
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