Atoms and Nuclei JEE Main Questions 2025: Important Practice Questions with PYQs

Atoms and Nuclei JEE Main Questions 2025: Atoms and Nuclei is an important topic from the JEE Mains Physics section. As per previous years' analysis, around 2 to 3 questions are asked from the topic which is approximately 8 to 12 marks in the JEE Mains exam. Some of the important topics asked from Atoms and Nuclei are Bohr’s Model of the Atom, Energy Level Transitions, Photoelectric Effect, de Broglie Hypothesis, Radioactivity, Binding Energy, and Nuclear Reactions. In this article, we have provided the important and solved Previous Years' Atoms and Nuclei JEE Main Questions along with some important questions for the JEE Mains 2025 Physics exam. 

Check Also: JEE Main Previous Year Question Paper

JEE Main 2025 Atoms and Nuclei Important Questions

Candidates can practice the Atoms and Nuclei JEE Main Questions based on the previous years from the questions below.

1. Consider the nuclear fission Ne²⁰ →2He⁴ + C¹²

Given that the binding energy/ nucleon of Ne20, He4 and C12 are 8.03 MeV, 7.07 MeV and 7.86 MeV respectively. Identify the correct statement.

(a) The energy of 9.72 MeV has to be supplied

(b) The energy of 12.4 MeV will be supplied

(c) 8.3 MeV energy will be released

(d) The energy of 3.6 MeV will be released

Solution

Ne²º →2He⁴ + C¹²

Q – value, EB = (BE)initial – (BE)final

= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12) = 9.72 MeV

Answer: (a) energy of 9.72 MeV has to be supplied

2. An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8: 27. The ratio of the radii of the nuclei (assumed to be spherical) is

(a) 3: 2

(b) 2: 3

(c) 4: 9

(d) 8: 27

Solution

As (v1/v2) = 8/27; (r1/r2) = ?

Using the law of conservation of linear momentum, 0 = m1v1 – m2v2

(As both are moving in opposite directions)

Or m1/m2 = v2/v1= 27/8

Therefore, (r1/r2) = 3/2

Answer: (a) 3: 2

3. According to Bohr’s theory, the time-averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to (n = principal quantum number)

(a) n–2

(b) n –3

(c) n–4

(d) n–5

Solution

Magnetic field at the centre, Bn = μ0I/2rn

For a hydrogen atom, the radius of the nth orbit is given by

rn = (n2/m) (h/2π)2(4πε0/e2)

rn ∝ n2

I = e/T = e/(2πrn/vn) = evn/2πrn

Also, vn ∝ n –1 ∴ I ∝ n –3

Hence, Bn ∝ n–5

Answer: (d) n–5

4. Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron = 1.1 MeV and helium = 7.0 MeV)

(a) 25.8 MeV

(b) 32.4 MeV

(c) 30.2 MeV

(d) 23.6 MeV

Solution

1H² + 1H² ➝ 2He⁴

Energy released = 4(Binding Energy of (1 2H)) – 4(Binding Energy of(2 4He)) = 4 × 7 – 4 × 1.1 = 23.6 MeV

Answer: (d) 23.6 MeV

5. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion

(a) kinetic energy decreases, potential energy increases but the total energy remains the same

(b) kinetic energy and total energy decrease but potential energy increases

(c) its kinetic energy increases but potential energy and total energy decrease

(d) kinetic energy, potential energy and total energy decrease.

Solution

For an electron in the nth excited state of the hydrogen atom,

kinetic energy = e2/(8πє0n2a0)

potential energy = – e2/(4πє0n2a0) and total energy = – e2/(8πє0n2a0)

where a0 is Bohr radius.

As an electron makes a transition from an excited state to the ground state, n decreases. Therefore kinetic energy increases but potential energy and total energy decrease.

Answer: (c) its kinetic energy increases but potential energy and total energy decrease

6. Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is

(a) 12.1 eV

(b) 36.3 eV

(c) 108.8 eV

(d) 122.4 eV

Solution

Using, En = -(13.6Z2/n2) eV

Here, Z = 3 (For Li++)

Therefore, E1 = -(13.6(3)2/12) = – 122.4 eV

and E3 =- (13.6(3)2/32) = – 13.6 eV

ΔE = E3 – E1 = – 13.6 + 122.4 = 108.8 eV

Answer: (c) 108.8 eV

7. If the binding energy per nucleon in 3Li⁷and 2He⁴ nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction

p + 3Li7 → 2 2He4

the energy of the proton must be

(a) 39.2 MeV

(b) 28.24 MeV

(c) 17.28 MeV

(d) 1.46 MeV

Solution

Binding energy of 3Li⁷ = 7 x 5.60 = 39.2 MeV

Binding energy of 2He⁴ = 4 x 7.06 = 28.24 MeV

Energy of proton = Energy of [2(2He⁴)- 3Li⁷]

= 2 × 28.24 – 39.2 = 17.28 MeV

Answer: (c) 17.28 MeV

8. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2: 1. The ratio of their nuclear sizes will be

(a) 21/3: 1

(b) 1: 31/2

(c) 31/2: 1

(d) 1: 21/3

Solution

Momentum is conserved during disintegration m1v1 = m2v2 ——-(1)

For an atom R = R0A1/3

(R1/R2) = (A1/A2)1/3 = (m1/m2)⅓ = (v2/v1)⅓ from (1)

(R1/R2) = (½)⅓ = 1/21/3

Answer: (d) 1: 21/3

9. A nuclear transformation is denoted by X(n,α)3Li⁷. Which of the following is the nucleus of element X?

(a) 5B⁹

(b) 4Be¹¹

(c) 6C¹²

(d) 5B¹º

Solution

The nuclear transformation is given by

zXA + 0n¹➝ 2He⁴ + 3Li⁷

According to the conservation of mass number

A + 1 = 4 + 7 or A = 10

According to the conservation of charge number

Z + 0 ➝ 2 + 3 or Z = 5

So the nucleus of the element is 5B¹º.

Answer: (d) 5B¹º

10. 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 at t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to

(a) 200

(b) 360

(c) 150

(d) 400

Solution

According to the law of radioactivity, the count rate at t = 8 seconds is

N1 = N0 e–λt

Taking logs on both sides

log N = log N0  – λt

λt = log N0 – log N

λt = log (N0/N )

At t = 0, the number of radioactive particles emitted is 1600

i.e., N0 = 1600

At t = 8 seconds,  the number of radioactive particles emitted is 100 i.e., N= 100

λ (8) = log (1600/100)

λ (8) = log 16

λ (8) = log 24

λ (8) = 4log 2

λ  = (log 2) /2

We have to find the number of radioactive particles emitting at t = 6 seconds. Therefore,

λt = log (N0/N )

[(log 2) /2] (6) = log (1600/N)

3 log 2 = log (1600/N)

log 23= log (1600/N)

log 8 = log (1600/N)

8 = (1600/N)

N = 200

Answer: (a) 200

11. Calculate the wavelength (in nanometers) associated with a proton moving at 1.0×103ms⁻¹ (Mass of proton = 1.67×10-27kg and h = 6.63×10-34Js)

(1) 2.5 nm

(2) 14.0 nm

(3) 0.032 nm

(4) 0.40 nm

Solution:

Given mp = 1.67×10-27kg

h = 6.63×10-34Js

v = 1.0×103ms⁻¹

We know wavelength λ = h/mv

∴λ = 6.63×10-34/(1.67×10-27 × 1.0×103)

= 3.97×10-10

≈ 0.04nm

Answer (4) 0.40 nm

12. If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become

(1) Two times

(2) Half

(3) One fourth

(4) Four times

Solution:

The wavelength λ is inversely proportional to the square root of kinetic energy. So if KE is increased 4 times, the wavelength becomes half.

λ∝1/√KE

Answer - (2) Half

Also Read: Best JEE Main Coaching Institutes in India

13. The radius of the second Bohr orbit for the hydrogen atom is :

(Planck’s constant, h = 6.262×10⁻³⁴Js: Mass of electron = 9.1091×10⁻³¹kg; Charge of electron e = 1.60210×10⁻¹⁹C; permittivity of vacuum ε0 = 8.854185×10⁻¹²kg-1m-3A²)

(1) 1.65 A

(2) 4.76 A

(3) 0.529 A

(4) 2.12 A

Solution:

The radius of nth Bohr orbit in H atom = 0.53 n2/Z

For hydrogen Z = 1

Radius of 2nd Bohr orbit in H atom = 0.53 ×22/1 = 2.12

Answer (4) 2.12 A

14. The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol-1. The longest wavelength of light capable of breaking a single Cl–Cl bond is

(C = 3×10⁸ m/s and NA = 6.02×10²³ mol⁻¹)

(1) 494 nm

(2) 594 nm

(3) 640 nm

(4) 700 nm

Solution:

We have B.E = 242KJ/Mol

E = hcNA/λ

∴ λ = hcNA/E

= 3×10⁸×6.626×10-34×6.02×10²³/(24²×10³)

= 0.494×10-3×10³

= 494 nm

Answer (1) 494 nm

15. The ionisation energy of He+ is 19.6×10⁻¹⁸J atom⁻¹. The energy of the first stationary state (n = 1) of Li²+ is

(1)8.82×10⁻¹⁶ J atom⁻¹

(2) 4.41×10⁻¹⁶ J atom⁻¹

(3) -4.41×10⁻¹⁷ J atom⁻¹

(4) -2.2×10⁻¹⁵ J atom⁻¹

Solution:

Given I.E = 19.6×10-18

I.E ∝ z²

(I.E) Li²+/He+ = (9/4)×19.6×10⁻¹⁸

= -4.41×10⁻¹⁷

Answer (3) -4.41×10⁻¹⁷ J atom⁻¹

15. The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in the H atom corresponding to which of the following

(1) n = 3 to n = 1

(2) n = 2 to n = 1

(3) n = 3 to n = 2

(4) n = 4 to n = 3

Solution:

E = 13.6×4[(¼)-(1/16)]

= 10.2

E = hν

ν = 10.2/h

E = 13.6(1)[(1/n12-1/n22)]

10.2 = 13.6[(1/n12-1/n22)]

102/136 = (n22-n12)/n12n22

Substitute the given options and find n1 and n2

51/68 = (n22-n12)/n12n22

0.75 = (4-1)4 = ¾ = 0.75

Answer (2) n = 2 to n = 1

16. Based on the equation ΔE = -2.0×10⁻¹⁸ J (1/n2 2⁻¹/n12) the wavelength of the light that must be absorbed to excite hydrogen electron from level n = 1 to level n= 2 will be (h = 6.625×10⁻³⁴ Js, C = 3×108 ms⁻¹)

(1) 2.650×10⁻⁷m

(2) 1.325×10⁻⁷m

(3) 1.325×10⁻¹ºm

(4) 5.300×10⁻¹ºm

Solution:

ΔE = -2.0×10⁻¹⁸ J (1/n2 2– 1/n12)

= -2.0×10⁻¹⁸(1/22 – 1/12)

= -2.0×10⁻¹⁸(1/4 – 1/1)

= -2.0×10⁻¹⁸(-3/4)

= 1.5×10⁻¹⁸

Also ΔE = hc/λ

So λ = hc/ΔE

= 6.625×10-34 × 3×108 /1.5×10⁻¹⁸

=13.25×10⁻⁸

= 1.325×10⁻⁷m

Answer (2) 1.325×10⁻⁷m

17. The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is :

(h = 6.63×10⁻³⁴ Js)

(1) 6.626×10⁻³¹ m

(2) 6.626×10⁻³⁴ m

(3) 6.626×10⁻³⁸ m

(4) 6.626×10⁻³º m

Solution:

Given h = 6.63×10⁻³⁴ J/s

m = 1000 kg

v = 36 km/hr = 36×103/(60×60) m/s = 10m/s

λ = h/mv

= 6.63×10⁻³⁴ /1000×10

= 6.63×10-38 m

Answer (3) 6.626×10⁻³⁸ m

18. If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is

(1) 13.6 eV 

(2) 30.6 eV 

(3) 122.4 eV 

(4) 3.4 eV

Solution:

B.E = 13.6×Z2/n2, Z is the atomic number and n is the orbital quantum number. For Li++, Z = 3 and n = 2 for the first excited state.

B.E = 13.6×32/22

= 30.6 ev

Answer (2) 30.6 eV 

19. According to Bohr’s theory, the angular momentum of an electron in 5th order orbit is

(1) 25 h/π

(2) 1.0 h/π

(3) 10 h/π

(4) 2.5 h/π

Solution:

n = 5

So angular momentum, = nh/2π

= 5h/2π

= 2.5 h/π

Answer (4) 2.5 h/π

20. The de Broglie wavelength of a tennis ball of mass 60g moving with a velocity of 10m/s is approximately ( Planck’s constant, h = 6.63×10⁻³⁴ Js)

(1) 10⁻³¹ m

(2) 10⁻¹⁶ m

(3) 10⁻²⁵ m

(4) 10⁻³³ m

Solution:

Given m = 60 g

v = 10 m/s

λ = h/mv

= 6.6×10⁻³⁴/(60×10⁻³×10) = 10⁻³³ m

Answer (4) 10⁻³³ m

21. In a hydrogen atom, if the energy of an electron in the ground state is 13.6 eV, then that in the 2nd excited state is

(1) 1.51 eV

(2) 3.4 eV

(3) 6.04 eV

(4)13.6 eV

Solution:

The 3rd energy level is the 2nd excited state.

n=3

En = 13.6/n² = 13.6/9 = 1.5 eV

Answer (1) 1.51 eV

22. Which of the following sets of quantum numbers is correct for an electron present in 4f orbital?

(1) n = 4, l = 3, m = +4, s = +½

(2) n = 3, l = 2, m = -2, s = +½

(3) n = 4, l = 3, m = +1, s = +½

(4) n = 4, l = 4, m = -4, s = -½

Solution:

For 4f orbital, n = 4 and l = 3.

Values of m = -3, -2, -1, 0, +1, +2, +3

Answer (3) n = 4, l = 3, m = +1, s = +½

23. The number of d-electrons retained in Fe²+ (At.no. of Fe = 26) ion is

(1) 4

(2) 5

(3) 6

(4) 3

Solution:

Configuration of Fe²+ = 3d⁶ 4s⁰

Answer (3) 6

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