JEE Main Previous Year Questions (2016- 2024): Modern Physics- 1 PDF Download

Q.1. Consider the following radioactive decay process

The mass number and the atomic number of A₆ are given by:      (JEE Main 2023)
(1) 210 and 84
(2) 210 and 82
(3) 211 and 80
(4) 210 and 80

Ans. (4)

Q.2. From the photoelectric effect experiment, following observations are made. Identify which of these are correct.
A. The stopping potential depends only on the work function of the metal.
B. The saturation current increases as the intensity of incident light increases.
C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light.
D. Photoelectric effect can be explained using wave theory of light.
Choose the correct answer from the options given below:        (JEE Main 2023)
(1) A, C, D only
(2) B, C only
(3) B only
(4) A, B, D only

Ans. (3)

Photoelectric effect is not explained by wave theory

Q.3. Assume that protons and neutrons have equal masses. Mass of a nucleon is 1.6 × 10⁻²⁷ kg and radius of nucleus is 1.5 × 10⁻¹⁵ A^1/3 m. The approximate ratio of the nuclear density and water density is n × 10¹³. The value of n is        (JEE Main 2023)

Ans. 11

Q.4. A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 × 10−15eVs ) :        (JEE Main 2023)
(1) 99.3 nm
(2) 941 nm
(3) 974 nm
(4) 94.1 nm

Ans. (4)

Q.5. An a-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:       (JEE Main 2023)
(1) ⁡λ_α < λ_p < λ_e
(2) λ_α = λ_p = λ_e
(3) λ_α > λ_p > λ_e
(4) λ_α > λ_p < λ_e

Ans. (1)

Q.6. The energy released per fission of nucleus of ⁡²⁴⁰ X is 200MeV. The energy released if all the atoms in 120 g of pure ⁡240X undergo fission is ______ × 10²⁵MeV (Given N_A = 6 × 10²³)        (JEE Main 2023)

Ans. 6

Q.7. A beam of electromagnetic radiation of intensity 6.4 x 10^-5 W/cm² is comprised of wavelength, λ = 310 nm. It falls normally on a metal (work function ϕ = 2eV) of surface area 1 cm² . If one in 10³ protons ejects an electron, total number of electrons ejected in 1s is 10^x,(hc = 1240 eV nm, 1eV 1.6 10^-19 J) then x is ________.    (2020)

Ans. (11)
Given that λ = 310 nm, hc = 1240 eV nm We know that photons will eject electrons if the energy of incident beam is greater than work function, that is, E > ϕ
⇒ hf > ϕ ⇒ hc/λ > ϕ
⇒
⇒ 4 eV > 2 eV
So, electrons will be ejected.
Now number of electrons ejected in 1s is
n_e =
Number of photon incident is given by

So,
Therefore, x = 11

Q.8. An electron (mass m) moving with initial velocityis in an electric fieldIf λ₀ is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by    (2020)
(1)
(2)
(3)
(4) 

Ans. (3)
Given that

de-Broglie wavelength before entering in electric filed is given by

Velocity of electron after entering electric filed at time t is given by


Since, electric filed is at z direction so it effects only z direction velocity.
So, de-Broglie wavelength after entering in electric filed at time t is given by 

=

Q.9. Radiation with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10⁻⁴ T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to    (2020)
(1) 1.1 eV
(2) 0.8 eV
(3) 1.6 eV
(4) 1.8 eV

Ans. (1)
Given that

Einstein’s Photoelectric Equation is given as
............ (1)
Radius of circular path of charged particle is magnetic field is given by

For maximum radius K should be maximum. So,
............. (2)
From Eqs. (1) and (2), we get

=
= 1.1eV

Q.10. A particle moving with kinetic energy E has de Broglie wavelength λ. If energy ΔE is added to its energy, the wavelength become λ/2. Value of ΔE, is    (2020)
(1) E
(2) 4E
(3) 3E
(4) 2E

Ans. (3)
de Broglie wavelength is given by


So,

Q.11. A electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is    (2020)
(a)
(b)
(c)
(d)

Ans. (4)
Electrostatic Force on a charge particle in a electric filed is given by
F = |e| E

Initially particle is at rest, that is, u = 0. So, velocity of particle after time t in electric filed is
De-Broglie wavelength is given by


So, the rate of change of de-Broglie wavelength of this electron at time t is given by

Q.12. The activity of a radioactive sample falls from 700/s to 500/s in 30 mins. Its half-life is close to    (2020)
(1) 72 min
(2) 62 min
(3) 66 min
(4) 52 min

Ans. (2)
Given that, A₀ = 700 /s, A = 500 /s and t = 30 min.
We know that, radioactivity at instant t is given by

We know that, half-life is given by

Q.13. The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10⁻¹⁶s. The frequency of revolution of the electron in its first excited state (in/s) is    (2020)
(1) 1.6 × 10¹⁴
(2) 7.8 × 10¹⁴
(3) 6.2 × 10¹⁵
(4) 5.6 × 10¹²

Ans. (2)
Given that T =
Time period of revolution of n^th orbit of radius r is given by
here
So,


For hydrogen atom Z = 1, T ∝ n³ 
We have n = 1, T = 1.6 × 10^-16s
For n = 2, we have

Therefore, frequency is

Q.14. The graph which depicts the results of Rutherford's gold foil experiment with α-particles is:
[θ: Scattering angle
Y: Number of scattered α-particles detected]
(Plots are schematic and not to scale)    (2020)
(1)
(2)
(3)
(4)

Ans. (4)
Relation between scattering angle (θ) and number of scattered α-particle (Y) detected is given by
 
OR

So, plot of option (4) depicts the correct relation between scattering angle and number of scattered α particle detected.

Q.15. The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is __________.    (2020)

Ans. (486)
Given that
For Balmer series, wavelength is given by
For first member, we have
................. (i)
For second member, we have
................... (ii)
From Eqs. (1) and (2), we have



= 486nm

Q.16. The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?    (2020)
(1) 24.2 nm
(2) 11.4 nm
(3) 35.8 nm
(4) 8.6 nm

Ans. (2)
Given that E_I = 9Rydbergs
We know that 1Rydberg = 13.6eV
Ionization energy is given by E_I = 13.6 eV
So, 13.6Z² = 9 x 13.6
Z² = 9
Z = 3
The wavelength of the radiation emitted when the electron in this ion jumps from the
second excited state to the ground state is given by

Q.17. Surface of certain metal is first illuminated with light of wavelength λ₁ = 350 nm and then, by light of wavelength λ₂ = 540 nm. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to    (2019)
(1) 1.8
(2) 2.5
(3) 5.6
(4) 1.4

Ans. (1)

Now, dividing Eq. (1) by Eq. (2), we get

Q.18. The magnetic field associated with a light wave is given, at the origin, by B = B₀[sin (3.14 × 10⁷)ct + sin (6.28 × 10⁷)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photoelectrons?
(c = 3 × 10⁸ m/s, h = 6.6 × 10⁻³⁴ J s)    (2019)
(1) 6.82 eV
(2) 12.5 eV
(3) 8.52 eV
(4) 7.72 eV

Ans. (4)
Given, magnetic field associated with light wave is
B = B₀ [sin(3.14 x 10⁷ c)t + sin(6.28 x 10⁷ c)t]    ...(1)
where c is the speed of light
In above wave equation, there are two electromagnetic waves with different frequency.
To get maximum kinetic energy consider the photon with higher frequency
B₁ = B₀sin(π × 10⁷ c)t

B₂ = B₀sin(2π × 10⁷ c)t
v₂ = 10⁷c
v₂ > v₁ 
Kinetic energy of photoelectron will be a maximum for photon of higher energy.

E_Ph = hf = 6.6 × 10⁻³⁴ × 10⁷ × 3 × 10⁶ 
= 6.6 × 3 × 10⁻¹⁹

Q.19. In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10⁻¹² m, the minimum electron energy required is close to    (2019)
(1) 500 keV
(2) 100 keV
(3) 1 keV
(4) 25 keV

Ans. (4)
Wavelength is given by

Therefore, required energy is

Q.20. In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to     (2019)
(1) 0.5 V
(2) 1.5 V
(3) 1.0 V
(4) 2.0 V

Ans. (3)
The potential necessary to stop any electron from reaching the other side
λ₁ = 300 nm
λ₂ = 400 nm
For λ₁ 

Subtracting Eq. (2) from Eq. (1), we get

Q.21. In a Frank–Hertz experiment, an electron of energy 5.6 eV passes through mercury vapor and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to    (2019)
(1) 1700 nm
(2) 2020 nm
(3) 220 nm
(4) 250 nm

Ans. (4)
In Frank–Hertz experiment

Q.22. When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photocurrent is -V₀/2 . When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is -V₀. The threshold frequency for photoelectric emission is    (2019)
(1) 5v/3
(2) 4v/3
(3) 2v
(4) 3v/2

Ans. (4)
Energy of incident photon is

From Eq. (1) and Eq. (2), we get

Q.23. A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 Å. The radius of the atom in the excited state, in terms of Bohr radius a₀, will be (hc = 12,500 eV Å)    (2019)
(1) 25 a₀ 
(2) 9 a₀ 
(3) 4 a₀ 
(4) 16 a₀ 

Ans. (4)
Energy of Photon is

Since, electron will excite to n = 4

Q.24. If the de Broglie wavelength of an electron is equal to 10⁻³ times the wavelength of a photon of frequency 6 × 10¹⁴ Hz, then the speed of electron is equal to (Given: speed of light = 3 × 10⁸ m/s; Planck’s constant = 6.63 × 10⁻³⁴ J s; mass of electron = 9.1 × 10⁻³¹ kg)    (2019)
(1) 1.45 × 10⁶ m/s
(2) 1.1 × 10⁶ m/s
(3) 1.7 × 10⁶ m/s
(4) 1.8 × 10⁶ m/s

Ans. (1)
We have,

Q.25. In a hydrogen-like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is λ. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be    (2019)

Ans. (4)
From M-shell to L-shell

From N-shell to L-shell

Dividing Eq. (1) by Eq. (2), we get

Q.26. A particle of mass m moves in a circular orbit in a central potential field If Bohr’s quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as    (2019)

Ans. (1)

In circular motion

Bohr quantization is

From Eq. (1), we get

From Eq. (1), we get

Q.27. A particle A of mass m and charge q is accelerated by a potential difference of 50 V. Another particle B of mass 4 m and charge q is accelerated by a potential difference of 2500 V. The ratio of de Broglie wavelengths l_A /l_B is close to    (2019)
(1) 10.00
(2) 0.07
(3) 14.14
(4) 4.47

Ans. (3)
Wave length is given by

Q.28. A sample of radioactive material A, that has an activity of 10 m Ci (1 Ci = 3.7 × 10¹⁰ decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 m Ci. The correct choices for half-lives of A and B would then be, respectively,    (2019)
(1) 5 days and 10 days.
(2) 10 days and 40 days.
(3) 20 days and 5 days.
(4) 20 days and 10 days.

Ans. (3)
Activity is given as,
A = λN
for A  λ_A N_A = 10   (1)
for B  λ_B N_B = 20   (2)
Dividing Eq. (1) by Eq. (2), we get

(T_1/2)A = 20 days
(T_1/2)B = 5 days

Q.29. At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio R_B/R_A of their activities after time t itself decays with time t as e^−3t. If the half-life of A is ln 2, the half-life of B is    (2019)
(1) 4ln2
(2) ln2/2
(3) ln2/4
(4) 2ln2

Ans. (3)
We know that,
N = N₀e^−λt 
If T_1/2 be the half-life period, then at t = T_1/2 and N = N₀/2

Half-life of A = ln 2    [ln = log_e]
λ_A = 1
At t = 0      R_A = R_B 

Q.30. Using a nuclear counter, the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to    (2019)
(1) 200
(2) 150
(3) 400
(4) 360

Ans. (1)
At t = 0 s

Q.31. Consider the nuclear fission Ne²⁰ → 2He⁴ + C¹². Given that the binding energy/nucleon of Ne²⁰, He⁴ and C¹² are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:    (2019)
(1) Energy of 12.4 MeV will be supplied.
(2) 8.3 MeV energy will be released.
(3) Energy of 3.6 MeV will be released.
(4) Energy of 11.9 MeV has to be supplied.

Ans. (4)
Ne² → 2He⁴ + C¹² 
Amount of energy
ΔE = 2 × (Binding energy of He⁴) + (Binding energy of C¹⁰) – (Binding energy of Ne²⁰)
= 2 × (4 × 7.07) + (12 × 7.86) – (20 × 8.03)
= 56.56 + 94.32 – 160.6
= −9.72 MeV
Hence, the energy of 11.9 MeV has to be supplied.

Q.32. In a radioactive decay chain, the initial nucleus is  At the end there are 6 α-particles and 4 β-particles which are emitted. If the end nucleus is , A and Z are given by    (2019)
(1) A = 208; Z = 80
(2) A = 202; Z = 80
(3) A = 208; Z = 82
(4) A = 200; Z = 81

Ans. (3)
We have,

Hence Z = 82, A = 208 and element is Pb.

Q.33. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λ_n,λ_g be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let λ_n be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)    [2018]
(1)  
(2)
(3) 
(4)

Ans: (3)
2πr = nλ_n



From equation (1) and (2)

Q.34.  If the series limit frequency of the Lyman series is  then the series limit frequency of the Pfund series is:    [2018]
(1) 25 v_L
(2) 16 v_L
(3)

(4)

Ans: (4)
Series limit frequency of the Lyman series is given by

Series limit frequency of the Pfund series

Q.35. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively:    [2018]
(1) (.89, .28)
(2) (.28, .89)
(3) (0, 0)
(4) (0, 1)

Ans: (1)

 (from conservation of momentum)
and K₀ = K₁ + K₂ (for elastic collision)
So after solving

Q.36. Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ₁ and λ₂, their de Broglie wavelength in the frame of reference attached to their centre of mass is:     [2018]
(1) λ_CM = λ₁ = λ₂
(2)  

(3)  

(4)

Ans: (3)
Momentum of each electron  
Velocity of centre of mass

velocity of 1st particle about centre of mass