Atoms & Nuclei Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions for Atoms & Nuclei with step-by-step solutions covering Bohr model, binding energy, nuclear reactions and radioactive decay.
Solved JEE Main 2026 previous year questions on Atoms & Nuclei — spanning the Bohr model, hydrogen spectra, binding energy, nuclear fusion/fission and radioactive decay — each with a concise, worked solution.
Solutions are AI-generated and pending review.
Solution
Reaction: two nuclei of $A=3$ plus one nucleus of $A=4$ give one nucleus of $A=10$.
Step 1 — Total binding energy of reactants:
$$BE_{\text{react}} = 2(3 \times 5.6) + (4 \times 7.4) = 33.6 + 29.6 = 63.2\ \text{MeV}$$Step 2 — Binding energy of product:
$$BE_{\text{prod}} = 10 \times 6.1 = 61.0\ \text{MeV}$$Step 3 — Mass-energy change: The bound-mass difference corresponds to
$$\Delta M c^2 = \left| BE_{\text{prod}} - BE_{\text{react}} \right| = |61.0 - 63.2| = 2.2\ \text{MeV}$$Since the product is less tightly bound, the reaction is endothermic; the magnitude $\Delta M c^2 = 2.2$ MeV.
Answer: C
Solution
Photon momentum $p = E/c \propto \bar{\nu}$ (wavenumber). For the Balmer series ($n_f = 2$):
$$\bar{\nu} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right)$$1st line ($n = 3$):
$$\bar{\nu}_1 = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\cdot\frac{5}{36}$$2nd line ($n = 4$):
$$\bar{\nu}_2 = R\left(\frac{1}{4} - \frac{1}{16}\right) = R\cdot\frac{3}{16}$$Ratio:
$$\frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{5 \times 16}{36 \times 3} = \frac{80}{108} = \frac{20}{27}$$So $\alpha = 20,\ \beta = 27$.
Answer: D
Solution
Reaction: $_3^7\text{Li} + {}_1^1\text{p} \rightarrow 2\,{}_2^4\text{He}$
Step 1 — Mass defect per reaction:
$$\Delta m = (7.0183 + 1.008) - 2(4.004) = 8.0263 - 8.008 = 0.0183\ \text{u}$$Step 2 — Energy per reaction:
$$E_1 = 0.0183 \times 931 = 17.04\ \text{MeV} = 1.704 \times 10^{7}\ \text{eV}$$Step 3 — Number of Li nuclei in $\dfrac{7}{17.13}\ \text{kg} = 408.64\ \text{g}$:
$$N = \frac{408.64}{7.0183} \times 6.0 \times 10^{23} \approx 3.49 \times 10^{25}$$Step 4 — Total energy (one Li consumed per reaction):
$$E = N \times E_1 = 3.49 \times 10^{25} \times 1.704 \times 10^{7} \approx 5.95 \times 10^{32}\ \text{eV}$$So $\alpha \approx 5.95 \Rightarrow 6$ (nearest integer).
Answer: 6
Solution
Step 1 — Find $n$ from Bohr quantization:
$$L = \frac{nh}{2\pi} = \frac{3h}{\pi} \implies \frac{n}{2} = 3 \implies n = 6$$Step 2 — Energy of the level:
$$E_n = -\frac{13.6}{n^2}\ \text{eV} = -\frac{13.6}{36} = -0.378\ \text{eV} \approx -0.38\ \text{eV}$$Answer: C
Solution
Evaluate each statement:
- A — True. The nucleus occupies a tiny fraction of the atom, so most $\alpha$-particles pass undeflected and only a few come close enough to rebound.
- B — False. The gold nucleus has charge $+79e$; the $\alpha$-particle has $+2e$. Their charges are not equal.
- C — True. Large-angle (rebound) scattering occurs only for the few particles aimed nearly at the nucleus, i.e. with minimum impact parameter.
- D — False. The incident $\alpha$-particles all have essentially the same kinetic energy; rebound is not caused by a few having higher energy.
- E — True. Only the few particles undergoing near head-on collisions are turned back through large angles.
Correct statements: A, C, E.
Answer: D
Solution
For $^{209}_{83}\text{Bi}$: $Z = 83$ protons, $N = 209 - 83 = 126$ neutrons.
Step 1 — Mass defect (using $m_p$ = hydrogen-atom mass, electron masses cancel):
$$\Delta m = 83(1.007825) + 126(1.008665) - 208.980388$$$$\Delta m = 83.649475 + 127.09179 - 208.980388 = 1.760877\ \text{u}$$Step 2 — Total binding energy:
$$BE = 1.760877 \times 931 = 1639.4\ \text{MeV}$$Step 3 — Binding energy per nucleon:
$$\frac{BE}{A} = \frac{1639.4}{209} = 7.84\ \text{MeV}$$Answer: B
Solution
The electron orbiting at radius $r$ with speed $v$ acts as a current loop producing a field at the nucleus:
$$B = \frac{\mu_0 I}{2r}, \qquad I = \frac{e v}{2\pi r} \implies B = \frac{\mu_0 e v}{4\pi r^2}$$Step 1 — Bohr scaling ($v \propto 1/n$, $r \propto n^2$):
$$B \propto \frac{v}{r^2} \propto \frac{1/n}{n^4} = \frac{1}{n^5}$$Step 2 — Take the ratio for $n = 2$ and $n = 4$:
$$\frac{B_2}{B_4} = \frac{n_4^5}{n_2^5} = \left(\frac{4}{2}\right)^5 = 2^5 = 32$$Answer: 32
Solution
The given binding energy per nucleon for $^2_1\text{H}$ (deuterium) tells us the fusion is:
$$2\,{}^2_1\text{H} \rightarrow {}^4_2\text{He}$$Step 1 — Binding energy of reactants (two deuterons, 2 nucleons each):
$$BE_{\text{react}} = 2 \times (2 \times 1.1) = 4.4\ \text{MeV}$$Step 2 — Binding energy of product ($^4_2\text{He}$, 4 nucleons):
$$BE_{\text{prod}} = 4 \times 7.2 = 28.8\ \text{MeV}$$Step 3 — Energy released:
$$Q = BE_{\text{prod}} - BE_{\text{react}} = 28.8 - 4.4 = 24.4\ \text{MeV}$$Answer: B
Solution
Step 1 — Find the orbit numbers using $r \propto n^2$:
$$r_i \propto n_i^2 = 16 \implies n_i = 4, \qquad r_f \propto n_f^2 = 4 \implies n_f = 2$$Step 2 — Apply the Rydberg formula:
$$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) = 1.0973 \times 10^7\left(\frac{1}{4} - \frac{1}{16}\right)$$$$\frac{1}{\lambda} = 1.0973 \times 10^7 \times \frac{3}{16} = 2.057 \times 10^6\ \text{m}^{-1}$$Step 3 — Wavelength:
$$\lambda = \frac{1}{2.057 \times 10^6} = 4.86 \times 10^{-7}\ \text{m} = 486\ \text{nm}$$Answer: C
Solution
$^{12}_{6}\text{C}$ has $6$ protons and $6$ neutrons.
Step 1 — Mass defect:
$$\Delta m = 6\,m_p + 6\,m_n - m(^{12}_6\text{C})$$$$\Delta m = 6(1.00727) + 6(1.00866) - 12 = 6.04362 + 6.05196 - 12 = 0.09558\ \text{u}$$Step 2 — Convert to MeV/$c^2$:
$$\Delta m = 0.09558 \times 931.5 = 89.03\ \text{MeV}/c^2$$Answer: B
Solution
When a parent (mass number $A$) at rest emits an $\alpha$-particle ($A_\alpha = 4$), momentum conservation gives the $\alpha$-particle kinetic energy:
$$K_\alpha = Q \cdot \frac{A - 4}{A}$$Step 1 — For A ($A = 200$):
$$K_A = Q \cdot \frac{196}{200}$$Step 2 — For B ($A = 212$):
$$K_B = Q \cdot \frac{208}{212}$$Step 3 — Ratio (same $Q$):
$$\frac{K_A}{K_B} = \frac{196/200}{208/212} = \frac{196 \times 212}{200 \times 208} = \frac{41552}{41600} = \frac{2597}{2600}$$Answer: C