JEE Main · Physics · Chapter 23

Atoms and Nuclei JEE Main PYQs (2020–2026)

205 questions from Atoms and Nuclei have appeared in JEE Main between 2020 and 2026, including 163 in the last five papers (2022–2026). The most-asked topic is Hydrogen Spectrum and Spectral Transitions (48 questions). Practise them all with instant answers and step-by-step solutions.

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205

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Topics

Numerical

Solved Atoms and Nuclei PYQs

3 recent questions with answers. The full set of 205 is free to practise.

JEE Main 2026EasyNuclear Fission and Nuclear Fusion
Q1
The energy released if hydrogen atoms are combined to form $_{2}^{4} He$ is $____$ MeV .

(Take binding energies per nucleon of $_{1}^{2} H$ and $_{2}^{4} He$ as 1.1 MeV and 7.2 MeV , respectively)
- A6.1
- B24.4
- C26.6
- D5
Show answer & solution
Correct answer: (B)
The binding energy (B.E.) of a nucleus is the energy required to disassemble it into its constituent nucleons.

Total B.E. $= ($ Binding Energy per nucleon $) \times ($ Number of nucleons A $)$

Energy Released (Q) = (Total B.E. of Products) - (Total B.E. of Reactants)

The given reaction is $H_{1}^{2} + H_{1}^{2} \to He_{2}^{4}$

The reactants are two nuclei of $H_{1}^{2}$ .

B.E. per nucleon of $H_{1}^{2} = 1.1 M e V$

Number of nucleons (A) for one $H_{1}^{2} = 2$

So, total B.E. of reactants $= 2 \times (1.1 \times 2) = 4.4 M e V$

The product is one nucleus of $He_{2}^{4}$ .

B.E. per nucleon of $He_{2}^{4} = 7.2 M e V$

Number of nucleons (A) for $He_{2}^{4} = 4$

So, total B.E. of product $= 7.2 \times 4 = 28.8 M e V$

Hence, energy released during reaction is,

$Q = 28.8 M e V - 4.4 M e V$

$\Rightarrow Q = 24.4 M e V$

Therefore, the energy released if hydrogen atoms are combined to form $He_{2}^{4}$ is 24.4 MeV .

Thus, the correct option is (D).
JEE Main 2026HardHydrogen Spectrum and Spectral Transitions
Q2
The ratio of momentum of the photons of the $1^{st}$ and $2^{nd}$ line of Balmer series of Hydrogen atoms is $α / β$ . The possible values of $α$ and $β$ are:-
- A27 and 20
- B3 and 16
- C5 and 36
- D20 and 27
Show answer & solution
Correct answer: (D)
According to the de Broglie relation, the momentum (p) of a photon is inversely proportional to its wavelength ( $λ$ ):

$p = \frac{h}{λ}$

Therefore, the ratio of momenta is the inverse ratio of their wavelengths:

$\frac{p _{1}}{p _{2}} = \frac{λ _{2}}{λ _{1}}$

The Balmer series occurs when an electron transitions from a higher energy level $(n_{2})$ to the second energy level ( $n_{1} = 2$ ). The wavelength is given by Rydberg formula:

$\frac{1}{λ} = R (\frac{1}{2 ^{2}} - \frac{1}{n ^{2}})$

For $1^{st}$ line of Balmer transition from $n = 3$ to $n = 2$ .

$\frac{1}{λ _{1}} = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{9 - 4}{36}) = \frac{5 R}{36} \Rightarrow λ_{1} = \frac{36}{5 R}$

For $2^{n d}$ line of Balmer transition from $n = 4$ to $n = 2$ .

$\frac{1}{λ _{2}} = R (\frac{1}{4} - \frac{1}{16}) = R (\frac{4 - 1}{16}) = \frac{3 R}{16} \Rightarrow λ_{2} = \frac{16}{3 R}$

So, the ratio $(α / β)$ ,

$\frac{p _{1}}{p _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{16/3 R}{36/5 R} = \frac{16}{3} \times \frac{5}{36}$

$\Rightarrow \frac{p _{1}}{p _{2}} = \frac{4 \times 5}{3 \times 9} = \frac{20}{27} = \frac{α}{β}$

So, $α = 20$ and $β = 27$ .

Therefore, the correct option is (D).
JEE Main 2026MediumMass Defect and Binding Energy
Q3
Assuming the experimental mass of $_{6}^{12} C$ as $12 u$ , the mass defect of $_{6}^{12} C$ atom is $____M e V / c^{2}$ .

(Mass of proton $= 1.00727 u$ . mass of neutron $= 1.00866 u, 1 u = 931.5 M e V / c^{2}$ and c is the speed of the light in vacuum).
- A127.5
- B89.03
- C272.0
- D92.0
Show answer & solution
Correct answer: (B)
Mass defect $(Δ m)$ is the difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual experimental mass of the nucleus.

For the atom $C_{6}^{12}$ , atomic number is $Z = 6$ (Number of protons)

Mass number is $A = 12$

Number of neutrons $N = A - Z = 12 - 6 = 6$

So, mass of 6 protons $= 6 \times 1.00727 u = 6.04362 u$

And the mass of 6 neutrons $= 6 \times 1.00866 u = 6.05196 u$

So, total theoretical mass $= 6.04362 + 6.05196 = 12.09558 u$

It is given that the experimental mass of $C_{6}^{12} = 12 u$ :

Hence, the mass defect is

$Δ m = (Total theoretical mass) - (Experimental mass)$

$\Rightarrow Δ m = 12.09558 u - 12 u = 0.09558 u$

The energy equivalent of 1 u is given as $931.5 M e V / c^{2}$ .

$Δ m (in M e V / c^{2}) = 0.09558 \times 931.5$

$\Rightarrow Δ m \approx 89.03277 \approx 89.03 M e V / c^{\land} 2$

Therefore, the mass defect of the $C_{6}^{12}$ atom is approximately $89.03 M e V / c^{2}$ .

Thus, the correct option is (B).