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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 1-30 of 316 questions

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1 Medium
1 of 316
Understanding +4 | -1
The electrons are emitted in the photoelectric effect from a metal surface
View answer and explanation
Correct option: A
Only if the frequency of the incident radiation is above a certain threshold value
Option A is correct because the photoelectric effect occurs only when the incident radiation's frequency exceeds the threshold frequency of the metal.
2 Medium
2 of 316
Application +4 | -1
What is the de Broglie wavelength of a bullet of mass 0.04 kg travelling at the speed of 1 km/s?
View answer and explanation
Correct option: A
1.7×10−35  m1.7 \times {10^{ - 35}}\,\,m
Option A (1.7×10−35 m1.7 \times {10^{ - 35}}\,m) is correct because it is the de Broglie wavelength calculated using the formula λ=hmv\lambda = \frac{h}{mv}.
3 Medium
3 of 316
Application +4 | -1
Photoelectrons are emitted from a metal surface when UV light of wavelength λ=300 nm\lambda=300 \mathrm{~nm} is incident on it. The minimum negative potential required to stop the emission of electrons is 0.54 V0.54 \mathrm{~V}. Calculate:
(i) the energy of the incident photons
(ii) the maximum kinetic energy of the photoelectrons emitted.
View answer and explanation
Correct option: A
4.14 eV;0.54 eV4.14 \mathrm{~eV}; 0.54 \mathrm{~eV}
Option A (4.14 eV;0.54 eV4.14 \mathrm{~eV}; 0.54 \mathrm{~eV}) is correct because it correctly calculates the energy of the incident photons and the maximum kinetic energy of the photoelectrons.
4 Medium
4 of 316
Application +4 | -1
An electron is accelerated by a potential difference of V volt to acquire kinetic energy equal to maximum kinetic energy of ejected photoelectron from a metal of work function 3.2 eV, using a light of wavelength 2000 A∘ 2000\,\mathop \mathrm{A}\limits^ \circ \, Value of V is
View answer and explanation
Correct option: C
3 volt
Option C (3 volt) is correct because the potential difference is equal to the maximum kinetic energy of the ejected photoelectron divided by the electron charge.
5 Medium
5 of 316
Analyzing +4 | -1
Two monochromatic beams AA and BB of equal intensity II, hit a screen. The number of photons hitting the screen by beam AA is twice that by beam BB. Then, what can inference you about their frequencies?
View answer and explanation
Correct option: A
The frequency of beam BB is twice that of AA
Option A is correct because the frequency of beam B is twice that of A, as derived from the relationship between intensity, number of photons, and frequency.
6 Medium
6 of 316
Application +4 | -1
A 5 W source emits monochromatic light of wavelength 5000A∘5000\mathop A\limits^ \circ. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0m, the number of photoelectrons liberated will be reduced by a factor of
View answer and explanation
Correct option: A
4
Option A (4) is correct because the number of photoelectrons is inversely proportional to the square of the distance from the source.
7 Medium
7 of 316
Application +4 | -1
An electron in a hydrogen atom makes a transition from n=n1  to  n=n2n = {n_1}\,\,\mathrm{to}\,\,n = {n_2} state. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1   and   n2{n_1}\,\,\,\mathrm{and}\,\,\,{n_2} are
View answer and explanation
Correct option: A
n1=4  ;  n2=2{n_1} = 4\,\,;\,\,{n_2} = 2
Option A (n1=4 ; n2=2{n_1} = 4\,;\,{n_2} = 2) is correct because the time period is proportional to n3n^3, so the ratio of the time periods is equal to the cube of the ratio of the quantum numbers.
8 Medium
8 of 316
Application +4 | -1
A photon possess energy of 2.21×10−162.21 \times 10^{-16} joule. The de-Broglie wavelength associated with the photon is (in A0\stackrel{0}{A} )
View answer and explanation
Correct option: A
9
Option A (9) is correct because it is the de-Broglie wavelength calculated using the given energy of the photon.
9 Medium
9 of 316
Understanding +4 | -1
Energy required by an electron for electron emission can be supplied to a free electron by

I. hammering the metal surface

II. heating the metal surface

III. applying electric field

IV. applying magnetic field

Which of the above statements is/are correct ?
View answer and explanation
Correct option: D
II and III
Option D (II and III) is correct because heating the metal surface (thermionic emission) and applying an electric field (field emission) can supply energy for electron emission.
10 Medium
10 of 316
Application +4 | -1
Ultraviolet light of wavelength 300 nm and intensity 1.0  Wm−21.0\;W{m^{ - 2}} falls on the surface on photoelectric metal. If one percent of incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0  cm21.0\;c{m^2} of the surface is nearly
View answer and explanation
Correct option: B
1.5×1012  s−11.5 \times {10^{12}}\;{\mathrm{s}^{ - 1}}
Option B (1.5×1012  s−11.5 \times {10^{12}}\;{\mathrm{s}^{ - 1}}) is correct because it calculates the number of photoelectrons emitted per second based on the given wavelength, intensity, and percentage of photon-electron conversion.
11 Medium
11 of 316
Remembering +4 | -1
Photo cell is a device to
View answer and explanation
Correct option: B
Measure light intensity
Option B (Measure light intensity) is correct because a photocell's primary function is to measure the intensity of light by converting photons into electrical signals.
12 Easy
12 of 316
Understanding +4 | -1
Which of the following phenomena does not explain by wave nature of light.

(A) reflection \quad(B) diffraction

(C) photoelectric effect\quad (D) interference\quad(E) polarization

Choose the most appropriate answer from the options given below :
View answer and explanation
Correct option: B
C only
Option B (C only) is correct because the photoelectric effect demonstrates the particle nature of light, not its wave nature.
13 Medium
13 of 316
Understanding +4 | -1
Particle nature and wave nature of electromagnetic waves and electrons can be shown by
View answer and explanation
Correct option: D
Photoelectricity and electron microscopy
Option D is correct because photoelectric effect demonstrates the particle nature of light, and electron microscopy uses the wave nature of electrons.
14 Medium
14 of 316
Understanding +4 | -1
A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become
View answer and explanation
Correct option: D
unchanged
Option D (unchanged) is correct because the maximum kinetic energy of emitted photoelectrons depends only on the frequency of the incident light, not its intensity.
15 Medium
15 of 316
Application +4 | -1
An electron of mass m and a photon have same energy E.E. The ratio of wavelength of electron to that of photon is (c being the velocity of light)
View answer and explanation
Correct option: B
1c(E2m)1/2\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{1/2}}
Option B (1c(E2m)1/2\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{1/2}}) is correct because it is derived from the ratio of the de Broglie wavelength of the electron to the wavelength of the photon.

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16 Medium
16 of 316
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.
17 Medium
17 of 316
Application +4 | -1
Two metallic plates AA and B,B, each of area 5×10−4m25 \times {10^{ - 4}}{m^2} are placed parallel to each other at a separation of 1  cm.1\;cm. plate BB carries a positive charge of 33.7  pc.33.7\;pc. A monochromatic beam of light. With photons of energy 5  eV5\;eV each, starts falling on plate AA at t=0,t = 0, so that 1016{10^{16}} photons fall on it per square meter per second. Assume that one photoelectron is emitted for energy 1016{10^{16}} incident photons. Also assume that all the emitted photoelectrons are collected by plate BB and the work function of plate AA remains constant at the value 2  eV.2\;eV. Electric field between the plates at the end of 10 seconds is
View answer and explanation
Correct option: A
2×103N/C2 \times {10^3}N/C
Option A (2×103N/C2 \times {10^3}N/C) is correct because it is the electric field calculated considering the photoelectrons emitted and collected.
18 Medium
18 of 316
Application +4 | -1
An α\alpha-particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de-Broglie wavelengths are λα{\lambda _\alpha } and λp{\lambda _p}, respectively. The ratio λαλp\frac{{{\lambda _\alpha }}}{{{\lambda _p}}} is
View answer and explanation
Correct option: D
2.8
Option D (2.8) is correct because the de-Broglie wavelength is inversely proportional to the square root of the product of mass and charge.
19 Medium
19 of 316
Application +4 | -1
Light of wavelength 500 nm is incident on a metal surface with work function 2.28 eV. Maximum kinetic energy of emitted electron is
View answer and explanation
Correct option: C
0.2 eV
Option D (1.44×109Nm−21.44 \times {10^9}\mathrm{N}{m^{ - 2}}) is correct because it is the result of calculating the bulk modulus using the given fractional compression and pressure at the specified depth.
20 Medium
20 of 316
Application +4 | -1
An alpha particle and a proton have same de - Broglie wave length. The ratio of their speed is
View answer and explanation
Correct option: C
0.044444444
Option C (01:04) is correct because the de Broglie wavelength is inversely proportional to momentum, and the alpha particle has 4 times the mass of a proton.
21 Medium
21 of 316
Application +4 | -1
Calculate the momentum of a photon of light of wavelength 500 nm500 \mathrm{~nm}.
View answer and explanation
Correct option: A
1.33×10−27 kg m s−11.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
Option A (1.33×10−27 kg m s−11.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}) is correct because it is calculated using the formula p=h/λp = h/\lambda where h is Planck's constant and λ\lambda is the wavelength.
22 Medium
22 of 316
Application +4 | -1
What is the momentum of a photon having frequency 1.5×1013Hz1.5 \times {10^{13}}Hz
View answer and explanation
Correct option: A
3.3×10−29kg  m/s3.3 \times {10^{ - 29}}kg\;m/s
Option A (3.3×10−29kg;m/s3.3 \times {10^{ - 29}}kg\\;m/s) is correct because it is the result of applying the formula for the momentum of a photon, p=hvcp = \frac{{hv}}{c}.
23 Medium
23 of 316
Understanding +4 | -1
If VV be the accelerating voltage, then the maximum frequency of continuous XX-rays is given by
View answer and explanation
Correct option: C
eVh\frac{{eV}}{h}
Option C (eVh\frac{{eV}}{h}) is correct because the maximum frequency of continuous X-rays is directly proportional to the accelerating voltage V and inversely proportional to Planck's constant h.
24 Medium
24 of 316
Application +4 | -1
Photoelectrons are emitted from a metal surface when UV light of wavelength λ=300 nm\lambda=300 \mathrm{~nm} is incident on it. The minimum negative potential required to stop the emission of electrons is 0.54 V0.54 \mathrm{~V}. Calculate the work function of the metal.
View answer and explanation
Correct option: A
3.6 eV3.6 \mathrm{~eV}
Option A (3.6 eV3.6 \mathrm{~eV}) is correct because it is calculated by subtracting the stopping potential energy from the incident photon energy.
25 Medium
25 of 316
Application +4 | -1
The work function of tungsten and sodium are 4.5eV4.5 {eV} and 2.3eV2.3 {eV} respectively. If the threshold wavelength, λ\lambda for sodium is 5460A∘5460 \mathop {\text{A}}\limits^ \circ, the value of λ\lambda for tungsten is
View answer and explanation
Correct option: A
2791A∘2791 \mathop {\text{A}}\limits^ \circ
Option A (2791A∘2791 \mathop {\text{A}}\limits^ \circ) is correct because it is calculated using the inverse relationship between work function and threshold wavelength.
26 Medium
26 of 316
Analyzing +4 | -1
Assertion :The de Broglie equation has significance for any microscopic or sub-microscopic particles.
Reason: The de Broglie wavelength is inversely proportional to the mass of the object if velocity is constant.
View answer and explanation
Correct option: A
If both Assertion & reason are True & the reason is a correct explanation of the assertion
Option A is correct because the de Broglie equation applies to microscopic particles, and the wavelength is inversely proportional to mass at constant velocity, explaining its significance.
27 Medium
27 of 316
Remembering +4 | -1
Momentum of a photon of wavelength λ\lambda is
View answer and explanation
Correct option: A
h/λh/\lambda
Option A (h/λh/\lambda) is correct because the momentum of a photon is given by the equation p=h/λp = h/\lambda, where h is Planck's constant and λ is the wavelength.
28 Medium
28 of 316
Application +4 | -1
Ultraviolet radiations of 6.2 eV falls on an aluminium surface. KE of fastest electron emitted is (work function = 4.2 eV)
View answer and explanation
Correct option: B
3.2×10−19J3.2 \times {10^{ - 19}}\mathrm{J}
Option B (3.2×10−19J3.2 \times {10^{ - 19}}\mathrm{J}) is correct because it is the kinetic energy calculated using the photoelectric effect equation.
29 Medium
29 of 316
Understanding +4 | -1
Assertion: Photoelectric effect is the phenomenon of emission of photon by metal when illuminated by light of suitable frequency.

Reason: An electron beam carries sufficient energy to release photons from the metal.
View answer and explanation
Correct option: D
Both Assertion and Reason are false
Option D is correct because both the assertion and the reason are false. The photoelectric effect involves the emission of electrons, not photons, and is caused by light, not an electron beam.
30 Medium
30 of 316
Understanding +4 | -1
Which of the following statement are true:

(a) Dual nature of matter is evident in de-Broglie equation λ=hp\lambda=\frac{\mathrm{h}}{\mathrm{p}}

(b) Wave nature of macroscopic particles not visible in daily life

(c) λ=hp\lambda=\frac{\mathrm{h}}{\mathrm{p}} is not satisfied by a photon
View answer and explanation
Correct option: B
(a, b)
Option B ((a, b)) is correct because the de-Broglie equation demonstrates the dual nature of matter, and the wave nature of macroscopic particles is not readily observable.
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