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JEE • PHYSICS • DUAL NATURE OF RADIATION AND MATTER

Dual nature of radiation and matter practice · Mixed difficulty

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Showing 271-300 of 316 questions

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271 Medium
271 of 316
Application +4 | -1
Intensity of U.V. radiations is 10 Watt/m210 \mathrm{~Watt} / \mathrm{m}^2 and its average wavelength is
6000Ao6000 \mathop {\text{A}}\limits^{\text{o}}. It is incident on a surface. Calculate number of photons per 2 sec. per 2 m22 \mathrm{~m}^2.
View answer and explanation
Correct option: A
1.2×10201.2 \times 10^{20}
Option A (1.2×10201.2 \times 10^{20}) is correct because it is the number of photons calculated using the given intensity, wavelength, and area over a time of 2 seconds.
272 Medium
272 of 316
Analyzing +4 | -1
Cut off potentials for a metal surface for light of wavelengths λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3 are V1,V2V_1, V_2 and V3V_3 respectively. If V1,V2V_1, V_2 and V3V_3 are in arithmatic progression, then λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3 are in
View answer and explanation
Correct option: C
Harmonic progression
Option C (Harmonic progression) is correct because the stopping potentials are in arithmetic progression, which implies the wavelengths are in harmonic progression.
273 Easy
273 of 316
Understanding +4 | -1
Statement-1: Photoelectric effect establishes quantum nature of light.
Statement-2: There is negligible time lag between photon collisions with the material and photoelectron emission irrespective of intensity of incident light. (Assume incident light is of frequency greater than threshold frequency of the material).
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1
Option A (Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1) is correct because the instantaneous emission of photoelectrons, regardless of intensity, supports the quantum nature of light.
274 Medium
274 of 316
Application +4 | -1
If wavelength of photon and electron is same then ratio of total energy of electron to total energy of photon would be
View answer and explanation
Correct option: A
Velocity  of  electron  Light′s  speed\frac{{\mathrm{Velocity\;of\;electron\;}}}{{\mathrm{Ligh}{\mathrm{t}^\mathrm{'}}\mathrm{s\;speed}}}
Option D (Velocity  of  electron  Light′s  speed\frac{{\mathrm{Velocity\;of\;electron\;}}}{{\mathrm{Ligh}{\mathrm{t}^{\mathrm{'}}}\mathrm{s\;speed}}}) is correct because the ratio of energies is proportional to the ratio of their velocities, given the same wavelength.
275 Medium
275 of 316
Application +4 | -1
Radius of central zone of circular zone plate is 2.3 mm. wavelength of incident light is 5893 A∘{A^ \circ }. Source is at a distance of 6 m. Then the distance of first image will be
View answer and explanation
Correct option: A
9 m
Option A (9 m) is correct because the focal length of the zone plate is calculated using the radius of the central zone and the wavelength of the incident light.
276 Medium
276 of 316
Analyzing +4 | -1
A proton and an electron have the same de Broglie wavelength. If Kp\mathrm{K}_{\mathrm{p}} and Ke\mathrm{K}_{\mathrm{e}} be the kinetic energies of proton and electron respectively. Then choose the correct relation:
View answer and explanation
Correct option: D
Kp<Ke\mathrm{K}_{\mathrm{p}}< \mathrm{K}_{\mathrm{e}}
Option D (Kp < Ke) is correct because since the de Broglie wavelengths are the same, the electron, having smaller mass, has a larger kinetic energy.
277 Medium
277 of 316
Analyzing +4 | -1
Assertion: Photoelectric effect is seen only when electron are bound.

Reason: Free electron cannot absorb whole energy of photon and consuming momentum and energy.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion
Option A is correct because the photoelectric effect requires bound electrons, and free electrons cannot conserve both energy and momentum when absorbing a photon.
278 Medium
278 of 316
Application +4 | -1
What will be the ratio of the de-Broglie wavelength of proton and α\alpha particle of same energy?
View answer and explanation
Correct option: A
2:12: 1
Option A (2:12: 1) is correct because the de-Broglie wavelength is inversely proportional to the square root of the mass when energy is constant.
279 Medium
279 of 316
Understanding +4 | -1
When the speed of electrons increase, then the value of its specific charge
View answer and explanation
Correct option: B
Decreases
Option B (Decreases) is correct because as the speed of electrons increases, their mass increases relativistically, causing the specific charge (e/m) to decrease.
280 Medium
280 of 316
Application +4 | -1
An α\alpha-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de-Broglie wavelengths are λα{\lambda _\alpha } and λp{\lambda _p} respectively. The ratio λpλα\frac{{{\lambda _p}}}{{{\lambda _\alpha }}}, to the nearest integer, is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the de-Broglie wavelength is inversely proportional to the square root of the product of charge and mass, and the ratio simplifies to approximately 3.
281 Medium
281 of 316
Analyzing +4 | -1
Consider the two following statements I and II, and identify the correct choice given in the answers

III. In photovoltaic cells the photoelectric current produced is not proportional to the intensity of incident light.

IV. In gas filled photoemissive cells, the velocity of photoelectrons depends on the wavelength of the incident radiation.
View answer and explanation
Correct option: D
I is false but II is true
Option D (I is false but II is true) is correct because the photoelectric current in photovoltaic cells is proportional to the intensity of incident light, and the velocity of photoelectrons in gas-filled photoemissive cells depends on the wavelength of incident radiation.
282 Medium
282 of 316
Remembering +4 | -1
G. P. Thomson experimentally confirmed the existence of matter waves by the phenomenon
View answer and explanation
Correct option: B
Refraction
Option A (Diffraction) is correct because G.P. Thomson experimentally confirmed the wave nature of electrons through diffraction experiments.
283 Medium
283 of 316
Understanding +4 | -1
Experimental study of photoelectric effect shows that

I. Photocurrent ∝\propto intensity of light.

II. saturation current ∝\propto intensity of light.

III. photoemission occurs only at frequency greater than threshold frequency.

IV. photoemission is an instantaneous process.

The correct statements are
View answer and explanation
Correct option: D
I, II, III and IV
Option D (I, II, III and IV) is correct because all the statements accurately describe the experimental observations of the photoelectric effect.
284 Medium
284 of 316
Application +4 | -1
When radiation of wavelength λ\lambda is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is
View answer and explanation
Correct option: B
4λ4\lambda
Option B (4λ4\lambda) is correct because the threshold wavelength is calculated using the photoelectric effect equations with two different wavelengths and stopping potentials.
285 Medium
285 of 316
Application +4 | -1
In Millikan’s oil drop experiment, a charged drop of mass 1.8×10−141.8 \times {10^{ - 14}} kg is stationary between the plates. The distance between the plates 0.9 cm and potential difference between the plates is 2000 V. The number of electrons on the oil drop is
View answer and explanation
Correct option: B
5
Option B (5) is correct because in Millikan's experiment, the number of electrons is calculated by equating the electric force to the gravitational force and solving for n.
286 Medium
286 of 316
Understanding +4 | -1
Statement 1: The photoelectric effect is proof of the quantised nature of light.

Statement 2: Each photon in light beam has same amount of energy
View answer and explanation
Correct option: A
Only statement 1 is correct
Option A (Only statement 1 is correct) is correct because the photoelectric effect demonstrates the quantized nature of light, but the energy of photons depends on wavelength.
287 Medium
287 of 316
Understanding +4 | -1
If a proton and electron have the same de Broglie wavelength, then
View answer and explanation
Correct option: D
Momentum of electron = momentum of proton
Option C (12I0\frac{1}{2}{I_0}) is correct because when unpolarized light passes through a polarizer, its intensity is reduced by half.
288 Medium
288 of 316
Remembering +4 | -1
Cathode rays are
View answer and explanation
Correct option: D
Electron waves
Option D (Electron waves) is correct because cathode rays are streams of electrons, which exhibit wave-particle duality.
289 Medium
289 of 316
Application +4 | -1
Threshold wavelength for photoelectric effect on sodium is 5000  A∘.5000\;\mathop A\limits^ \circ . Its work function is
View answer and explanation
Correct option: C
4×10−19J4 \times {10^{ - 19}}J
Option C (4×10−19J4 \times 10^{ - 19}J) is correct because the work function is calculated using the threshold wavelength and Planck's constant.
290 Medium
290 of 316
Application +4 | -1
The de Broglie wavelength of an electron having kinetic energy EE is λ\lambda. If the kinetic energy of electron becomes E4\frac{E}{4}, then its de-Broglie wavelength will be:
View answer and explanation
Correct option: D
2λ2 \lambda
Option D (2λ2 \lambda) is correct because the de Broglie wavelength is inversely proportional to the square root of the kinetic energy.
291 Medium
291 of 316
Application +4 | -1
If the work function of a metal is 6.63 eV6.63 \mathrm{~eV}, then find its threshold frequency for photoelectric effect.
View answer and explanation
Correct option: B
1.6×1015 Hz1.6 \times 10^{15} \mathrm{~Hz}
Option B (1.6×1015 Hz1.6 \times 10^{15} \mathrm{~Hz}) is correct because it is calculated directly from the work function using the formula f0=ϕ0hf_0 = \frac{\phi_0}{h}.
292 Medium
292 of 316
Remembering +4 | -1
α−\alpha - particle contain –
View answer and explanation
Correct option: C
2 proton, 2 neutron only
Option C (2 proton, 2 neutron only) is correct because an alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons.
293 Medium
293 of 316
Application +4 | -1
The number of photons per second on an average emitted by the source of monochromatic light of wave length 1200 nm, when it delivers the power 3.3×10−3  watt3.3 \times {10^{ - 3}}\,\,{\text{watt}} will be [h=6.6×10−34  J  s]\left[ {h = 6.6 \times {{10}^{ - 34}}\,\,J\,\,s} \right]
View answer and explanation
Correct option: B
2×10162 \times {10^{16}}
Option B (2×10162 \times {10^{16}}) is correct because the number of photons is calculated using the power, wavelength, Planck's constant, and the speed of light.
294 Medium
294 of 316
Application +4 | -1
Find the de-Broglie wavelength associated with an electron moving with a velocity 0.5c0.5c and rest mass =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg}.
View answer and explanation
Correct option: A
4.2×10−12 m4.2 \times 10^{-12} \mathrm{~m}
Option A (4.2×10−12 m4.2 \times 10^{-12} \mathrm{~m}) is correct because it is the de-Broglie wavelength calculated using the relativistic mass of the electron.
295 Medium
295 of 316
Understanding +4 | -1
The mean free path of the electrons in a discharge tube is 20 cm. The length of the tube is 15 cm only. Then length of Crooke’s dark space is
View answer and explanation
Correct option: C
15 cm
Option C (15 cm) is correct because the length of Crooke's dark space cannot exceed the length of the discharge tube.
296 Medium
296 of 316
Application +4 | -1
If wavelength of radiation is 4000Ao4000 \mathop {\text{A}}\limits^{\text{o}}, then the energy of the photon is.
View answer and explanation
Correct option: A
3.1 eV3.1 \mathrm{~eV}
Option A (3.1 eV3.1 \mathrm{~eV}) is correct because the energy of a photon is given by E=hcλE = \frac{{hc}}{\lambda}, and substituting the given wavelength yields approximately 3.1 eV.
297 Medium
297 of 316
Understanding +4 | -1
In photo electric effect

A. The photocurrent is proportional to the intensity
of the incident radiation.

B. Maximum kinetic energy with which
photoelectrons are emitted depends on the
intensity of incident light.

C. Max. K.E with which photoelectrons are emitted
depends on the frequency of incident light.

D. The emission of photoelectrons require a
minimum threshold intensity of incident
radiation.

E. Max. K.E of the photoelectrons is independent
of the frequency of the incident light.

Choose the correct answer from the options given
below:
View answer and explanation
Correct option: C
A and C only
Option C (A and C only) is correct because photocurrent is proportional to intensity, and maximum kinetic energy depends on the frequency of incident light.
298 Medium
298 of 316
Understanding +4 | -1
Assertion: In photo-particle collision the total energy and total momentum are conserved.

Reason: the number of photos are conserved in a collision.
View answer and explanation
Correct option: C
Assertion is true but Reason is false
Option C (Assertion is true but Reason is false) is correct because while energy and momentum are conserved in photon-particle collisions, the number of photons isn't necessarily conserved.
299 Medium
299 of 316
Understanding +4 | -1
Assertion: The kinetic energy of photoelectrons emitted from metal surface does not depend on the intensity of incident photon.

Reason: The ejection of electrons from metallic surface is not possible with frequency of incident photons below the threshold frequency.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B is correct because both the assertion and the reason are true, but the reason does not explain why the kinetic energy is independent of intensity.
300 Medium
300 of 316
Analyzing +4 | -1
Assertion (A): Work function of aluminium is 4.2 eV. Emission of electron will not be possible if two photons each of energy 2.5 eV strike an electron of a aluminium.
Reason (R): For photoelectron emission the energy of each photon should be greater than the work function of metal.
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A
Option A (Both A and R are true and R is the correct explanation of A) is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion.
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