Skip to main content
Back to DUAL NATURE OF RADIATION AND MATTER

Your Progress

0 / 316 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • DUAL NATURE OF RADIATION AND MATTER

Dual nature of radiation and matter practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 61-90 of 316 questions

Preparing practice and saved progress…

61 Medium
61 of 316
Understanding +4 | -1
Given below are tow statements:
Statement I The photons having equal linear momenta have equal wavelengths.
Statement II If the wavelength of photon is decreased, then the momentum and energy of a photon will and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: C
Statement I is true but statement II is false.
Statement I is true and Statement II is false. The correct answer is C because wavelength and momentum are inversely proportional, and decreasing wavelength increases momentum and energy.
62 Medium
62 of 316
Understanding +4 | -1
The maximum kinetic energy of emitted electrons in a photoelectric effect does not depend upon
View answer and explanation
Correct option: C
Intensity
Option C (Intensity) is correct because the maximum kinetic energy of emitted electrons in the photoelectric effect depends on the frequency of light and the work function of the metal, but not on the intensity of the light.
63 Medium
63 of 316
Understanding +4 | -1
An electromagnetic radiation has an energy of 13.2  keV.13.2\;keV. Then the radiation belongs to the region of
View answer and explanation
Correct option: D
X-ray
Option D (X-ray) is correct because electromagnetic radiation with an energy of 13.2 keV falls within the X-ray region of the electromagnetic spectrum.
64 Medium
64 of 316
Application +4 | -1
Find the de-Broglie wavelength for electrons which passed through a potential difference of 100 V ? [h=6.625×10−34J−sec⁡]\left[ {h = 6.625 \times {{10}^{ - 34}}J - \sec } \right]
View answer and explanation
Correct option: A
1.228A∘1.228\mathop \mathrm{A}\limits^ \circ
Option A (1.228A∘1.228\mathop \mathrm{A}\limits^ \circ) is correct because it is calculated using the de-Broglie wavelength formula.
65 Medium
65 of 316
Analyzing +4 | -1
Assertion: the stopping potential depends on the frequency of incident light.

Reason: the stopping potential is related to maximum kinetic energy by eV0=Kmax⁡e V_0=K_{\max }
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because the stopping potential is directly related to the maximum kinetic energy, which depends on the frequency of incident light.
66 Medium
66 of 316
Analyzing +4 | -1
Electric field and magnetic field in a region of space is given by E=E0j^\mathbf{E}=E_0 \hat{\mathbf{j}} and B=B0j^\mathbf{B}=B_0 \hat{\mathbf{j}}. A particle of specific charge α\alpha is released from origin with velocity v=v0i^\mathbf{v}=v_0 \hat{\mathbf{i}}. Then, path of particle
View answer and explanation
Correct option: C
is a helix with non-uniform pitch
Option C (is a helix with non-uniform pitch) is correct because the electric field accelerates the particle along the y-axis, while the magnetic field causes circular motion in the xz-plane, resulting in a helix with varying pitch.
67 Medium
67 of 316
Understanding +4 | -1
Consider a beam of electrons (each electron with energy E0E_0 ) incident on a metal surface kept in an evacuated chamber. Then,
View answer and explanation
Correct option: D
Electrons can be emitted with any energy, with a maximum of E0E_0
Option D is correct because electrons can be emitted with any energy up to the incident energy E0E_0 due to energy loss during interaction with the metal.
68 Medium
68 of 316
Remembering +4 | -1
A stream of fast moving negatively charged particles are called
View answer and explanation
Correct option: A
Cathode rays
Option A (Cathode rays) is correct because cathode rays are streams of fast-moving negatively charged particles (electrons).
69 Medium
69 of 316
Remembering +4 | -1
The energy of a photon of wavelength λ\lambda is given by
View answer and explanation
Correct option: D
hcλ\frac{h c}{\lambda}
Option D (hcλ\frac{{h c}}{{\lambda}}) is correct because the energy of a photon is given by E=hcλE = \frac{{h c}}{{\lambda}}.
70 Medium
70 of 316
Application +4 | -1
The sun radiates energy at the rate of 3.77×1026J/s.3.77 \times {10^{26}}J/s. The loss of mass it suffers per seconds is
View answer and explanation
Correct option: B
41.9×108kg41.9 \times {10^8}kg
Option B (41.9×108kg41.9 \times {10^8}kg) is correct because the mass loss is calculated using Einstein's mass-energy equivalence formula, E = mc².
71 Medium
71 of 316
Analyzing +4 | -1
Assertion: Among the particles of same kinetic energy; lighter particle has greater de Broglie - wavelength.

Reason: The de-Broglie wavelength of a particle depends only on the charge of the particle.
View answer and explanation
Correct option: C
If Assertions is true but the reason is false
The correct answer is C because the assertion is true (lighter particles have greater de Broglie wavelength at the same kinetic energy), but the reason is false (de Broglie wavelength depends on momentum, not just charge).
72 Medium
72 of 316
Understanding +4 | -1
The maximum kinetic energy of a photoelectron is 3eV3 {eV}. Its stopping potential is
View answer and explanation
Correct option: A
3V3 {V}
Option A (3 V) is correct because the stopping potential is numerically equal to the maximum kinetic energy in electron volts.
73 Medium
73 of 316
Application +4 | -1
The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV1.7 \mathrm{~eV} is 14.3 V14.3 \mathrm{~V}. Find the wavelength of radiation used:
View answer and explanation
Correct option: B
775 A∘775\,\mathop {\text{A}}\limits^ \circ
Option B (775 A∘775\,\mathop {\text{A}}\limits^ \circ) is correct because it's calculated using the photoelectric effect equation, relating stopping potential, work function, and wavelength.
74 Medium
74 of 316
Understanding +4 | -1
When an inert gas is filled in place of vacuum in a photo cell, then
View answer and explanation
Correct option: B
Photo-electric current is increased
Option A (Photo-electric current is decreased) is correct because the inert gas atoms can collide with photoelectrons, reducing their kinetic energy and probability of reaching the anode.
75 Medium
75 of 316
Analyzing +4 | -1
K1{K_1} be the maximum kinetic energy of photoelectrons emitted when metal surface is illuminated by light of wavelength λ1.{\lambda _1}. Similarly K2{K_2} is corresponding to wavelength λ2{\lambda _2} for same metal . If λ1=2λ2{\lambda _1} = 2{\lambda _2} then
View answer and explanation
Correct option: C
K1<K22{K_1} < \frac{{{K_2}}}{2}
Option C (K1<K22{K_1} < \frac{{{K_2}}}{2}) is correct because when the wavelength is halved, the kinetic energy more than doubles due to the work function.
76 Medium
76 of 316
Understanding +4 | -1
Assertion: Millikan's experiment establish that electric charge is quantized.

Reason: From this experiment mass of the electron could not be determined.
View answer and explanation
Correct option: C
Assertion is true but Reason is false
Option C is correct because Millikan's experiment did establish charge quantization, but it also allowed for the determination of the electron's mass.
77 Medium
77 of 316
Remembering +4 | -1
Which of the following is not the property of the photons?
View answer and explanation
Correct option: C
Charge
Option C (Charge) is correct because photons are electrically neutral particles, meaning they have no electric charge.
78 Medium
78 of 316
Analyzing +4 | -1
The de-Broglie wavelength of an electron, α\alpha- particle and proton all having the same kinetic energy is respectively given as λe,  λα    and  λp.{\lambda _e},\;{\lambda _\alpha }\;\;\mathrm{and\;}{\lambda _p}. Then which of the following is not true?
View answer and explanation
Correct option: A
λe<λp{\lambda _e} < {\lambda _p}
Option A is correct because the de-Broglie wavelength is inversely proportional to mass, and the electron's wavelength is greater than the proton's, not less.
79 Medium
79 of 316
Understanding +4 | -1
Statement 1: The energy (E)(E) and momentum (p) of a photon are related by p=E/cp=E / c.

Statement 2: The photon behaves like a particle.
View answer and explanation
Correct option: C
Both the statements are correct
Option C (Both the statements are correct) is correct because both the energy-momentum relationship for a photon and the particle-like behavior of photons are true.
80 Medium
80 of 316
Analyzing +4 | -1
While doing his experiment, Millikan one day observed the following charges on a single drop
(i)6.563×10−19C6.563 \times {10^{ - 19}}C (ii) 8.204×10−19C8.204 \times {10^{ - 19}}C
(iii)11.50  ×10−19C  11.50\; \times {10^{ - 19}}C\; (iv)13.13×10−19C13.13 \times {10^{ - 19}}C
(v)16.48×10−19C    16.48 \times {10^{ - 19}}C\;\; (vi)18.09×10−19C18.09 \times {10^{ - 19}}C
From this data the value of the elementary charge (e)\left( e \right) was found to be
View answer and explanation
Correct option: A
1.641×10−19C1.641 \times {10^{ - 19}}C
Option A (1.641×10−19C1.641 \times {10^{ - 19}}C) is correct because it is the elementary charge calculated from the given Millikan's oil drop experiment data.
81 Medium
81 of 316
Application +4 | -1
Light of frequency 1.5\mathrm{1}\mathrm{.5} times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled ?
View answer and explanation
Correct option: D
zero
Option D (zero) is correct because when the frequency is halved, it becomes less than the threshold frequency, resulting in no photoelectric emission and thus zero current.
82 Hard
82 of 316
Application +4 | -1
In a photoelectric experiment, the wavelength of the light incident on a metal in changed from 300 nm to 400 nm. The decrease in the stopping potential is close to (hce=1240 nmV)\left( {\frac{{hc}}{e} = 1240\,nmV} \right)
View answer and explanation
Correct option: D
1.0 V
Option D (1.0 V) is correct because the decrease in stopping potential is calculated using the change in wavelength and the given constant.
83 Medium
83 of 316
Understanding +4 | -1
In a photoelectric experiment, increasing the intensity of incident light
View answer and explanation
Correct option: D
increases the number of photons incident and the KE of the ejected electrons remains unchanged.
Option D is correct because increasing the intensity of incident light increases the number of photons, but the kinetic energy of ejected electrons depends only on the frequency of the light.
84 Medium
84 of 316
Application +4 | -1
Particle of mass ' MM ' at rest breaks into two particles of masses M3\frac{M}{3} and 2M3\frac{2 M}{3}. Find the ratio of their deBroglie wavelengths :
View answer and explanation
Correct option: B
0.042361111
Option B (01:01) is correct because the de Broglie wavelength is inversely proportional to momentum, and momentum is conserved in the break-up.
85 Medium
85 of 316
Understanding +4 | -1
A free neutron decays into a proton but a free proton does not decay into neutron. This is because
View answer and explanation
Correct option: A
Neutron has larger rest mass than proton
Option A is correct because the decay of a neutron into a proton requires energy, which is available since the neutron's rest mass is greater than the proton's.
86 Medium
86 of 316
Understanding +4 | -1
Statement 1: Kinetic energy of photo electrons emitted by a photosensitive surface depends upon the intensity of incident photon.

Statement 2: The ejection of electrons from metallic surface is possible with frequency of incident photon below the threshold frequency.
View answer and explanation
Correct option: D
Both the statements are wrong
Option D (Both the statements are wrong) is correct because the kinetic energy of photoelectrons depends on the frequency of incident light, not intensity, and photoelectric emission is not possible below the threshold frequency.
87 Medium
87 of 316
Understanding +4 | -1
Assertion: In photoelectric effect, at stopping potential emission of photoelectrons is stopped.

Reason: Increase in intensity of given light increases the kinetic energy of photoelectron.
View answer and explanation
Correct option: D
If both Assertion and reason are false
Option D (If both Assertion and reason are false) is correct because at stopping potential, photoelectrons are stopped, and increasing the intensity of light increases the number of photoelectrons, not their kinetic energy.
88 Medium
88 of 316
Analyzing +4 | -1
The photons of same frequencies moving in same medium have
View answer and explanation
Correct option: D
None of the above
Option D is correct because photons of the same frequency moving in the same medium have the same energies and speeds, but not necessarily the same linear momentum, as momentum is a vector quantity and depends on direction.
89 Medium
89 of 316
Understanding +4 | -1
Assertion: Every metal has a definite work function, still all photoelectrons do not come out with the same energy if incident radiation is monochromatic.
Reason: Work function is the minimum energy required for the electron in the highest level of the conduction band to get out of the metal. Not all electrons in the metal belong to this level rather they occupy a continuous band of levels.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
Option A (If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion) is correct because the reason accurately explains why photoelectrons have varying energies despite a definite work function.
90 Medium
90 of 316
Understanding +4 | -1
Which is the incorrect statement of the following
View answer and explanation
Correct option: B
Photon is a particle with zero momentum
The correct answer is B (Photon is a particle with zero momentum) because photons, while having zero rest mass, possess momentum given by p=Ec=hλp = \frac{E}{c} = \frac{h}{\lambda}.
Revision notes and key takeaways

Review DUAL NATURE OF RADIATION AND MATTER in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 316 questions

Your practice

0 / 316 answered · 316 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.