Atoms and Nuclei Previous Year Questions — JEE Main Physics
23 past-year Atoms and Nuclei questions from JEE Main (Physics) — page 4.
Q27 (2026)
7.9 MeV $\alpha$-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) \_\_\_\_ m.
$\left[\frac{1}{4\pi \epsilon_{0}} = 9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2} \text{ and electron charge } = 1.6 \times 10^{-19} \mathrm{C}\right]$
- $5.76 \times 10^{-14}$
- $1.44 \times 10^{-13}$
- $2.88 \times 10^{-14}$
- $1.69 \times 10^{-12}$
### Related Formula
$$r_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{K}$$
### Core Logic
{{SOL_IMG1}}
Using mechanical energy conservation:
$$PE_i + KE_i = PE_f + KE_f$$
$$0 + 7.9 \times 10^6 \times 1.6 \times 10^{-19} = \frac{k(2e)(Ze)}{r} + 0$$
$$r = \frac{9 \times 10^9 \times 2 \times (1.6 \times 10^{-19})^2 \times 79}{7.9 \times 10^6 \times 1.6 \times 10^{-19}} = 2.88 \times 10^{-14} \text{ m}$$
For diameter:
$$D = 2r = 5.76 \times 10^{-14} \text{ m}$$
### Pattern Recognition
Sees: Alpha particle scattering + atomic number + kinetic energy.
Shortcut: Equate initial kinetic energy to electrostatic potential energy at closest approach distance $r$, then double for diameter $D = 2r$.
Check: Results align with option (1). ✓
### Chapter Mix
Class 12 Physics: Atoms and Nuclei
Q36 (2026)
The minimum frequency of photon required to break a particle of mass 15.348 amu into 4$\alpha$ particles is \_\_\_\_ kHz.
[mass of He nucleus = 4.002 amu, 1 amu = $1.66 \times 10^{-27}$ kg, h = $6.6 \times 10^{-34}$ J.s and c = $3 \times 10^{8}$ m/s]
- $9 \times 10^{19}$
- $9 \times 10^{20}$
- $14.94 \times 10^{20}$
- $14.94 \times 10^{19}$
### Related Formula
$$h\nu = \Delta m \cdot c^2$$
### Core Logic
Mass defect calculation:
$$h\nu = (4 \times 4.002 - 15.348) \times 1.66 \times 10^{-27} \times (3 \times 10^{8})^{2}$$
$$\nu = 14.94 \times 10^{19} \text{ kHz}$$
### Pattern Recognition
Sees: Minimum photon frequency required for nuclear breakup.
Shortcut: Compute mass defect $\Delta m$, convert to energy via $E=\Delta m c^2$, and equate to $h\nu$.
Check: Matches option (4). ✓
### Chapter Mix
Class 12 Physics: Atoms and Nuclei
Q37 (2026)
Which of the following pair of nuclei are isobars of the element?
- ${}^{2}_{1}\mathrm{H} \text{ and } {}^{3}_{1}\mathrm{H}$
- ${}^{236}_{92}\mathrm{U} \text{ and } {}^{238}_{92}\mathrm{U}$
- ${}^{198}_{80}\mathrm{Hg} \text{ and } {}^{197}_{79}\mathrm{Au}$
- ${}^{3}_{1}\mathrm{H} \text{ and } {}^{3}_{2}\mathrm{He}$
### Related Formula
Definition: Isobars are nuclei that possess the same mass number (A) but different atomic numbers (Z).
### Core Logic
Checking the mass numbers for each pair:
(1) Hydrogen-2 and Hydrogen-3 -> Isotopes
(2) Uranium-236 and Uranium-238 -> Isotopes
(3) Mercury-198 and Gold-197 -> Neither isobars nor isotopes
(4) Hydrogen-3 (${}^3_1\mathrm{H}$) and Helium-3 (${}^3_2\mathrm{He}$) -> Both have mass number 3.
### Step 1: Final Conclusion
Since ${}^3_1\mathrm{H}$ and ${}^3_2\mathrm{He}$ share the same mass number, they are isobars.
### Pattern Recognition
Isobars = same top number. Isotopes = same bottom number. Isotones = same difference (neutrons).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q46 (2026)
The average energy released per fission for the nucleus of ${}^{235}_{92}\mathrm{U}$ is 190 MeV. When all the atoms of 47 g pure ${}^{235}_{92}\mathrm{U}$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ____.
(Avogadro Number = $6 \times 10^{23}$ per mole)
### Related Formula
$$\text{Number of Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}}$$
$$N = \text{Moles} \times N_A$$
$$E_{\text{total}} = N \times E_{\text{per\_fission}}$$
### Core Logic
Given mass of U-235 = $47 \, \mathrm{g}$.
Molar mass of U-235 = $235 \, \mathrm{g/mol}$.
Number of moles = $\frac{47}{235} = \frac{1}{5} \text{ moles}$.
### Step 1: Calculate Total Atoms
Total number of U-235 atoms ($N$):
$$N = \frac{1}{5} \times 6 \times 10^{23} = 1.2 \times 10^{23} \text{ atoms}$$
### Step 2: Calculate Total Energy
$$E_{\text{total}} = 1.2 \times 10^{23} \times 190 \, \mathrm{MeV}$$
$$E_{\text{total}} = 228 \times 10^{23} \, \mathrm{MeV}$$
Comparing with $\alpha \times 10^{23} \, \mathrm{MeV}$, we get $\alpha = 228$.
### Pattern Recognition
Direct stoichiometry-like energy yield. Mass / Molar Mass -> Atoms -> Total Energy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q38 (2026)
Given below are two statements:
Statement I : For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon.
Statement II : For all elements, nuclei with less binding energy per nucleon transforms to nuclei with greater binding energy per nucleon.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{Both Statement I and Statement II are true}$
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is false but Statement II is true}$
### Core Logic
Statement I: The binding energy per nucleon does not monotonically increase with mass. It reaches a maximum around iron ($A \approx 56$) and then gradually decreases for heavier nuclei. Hence, Statement I is false.
Statement II: Nuclei inherently tend toward states of greater stability. Thus, nuclei with lower binding energy per nucleon undergo reactions (fission or fusion) to form nuclei with higher binding energy per nucleon. Hence, Statement II is true.
### Step 1: Final Conclusion
Statement I is false but Statement II is true. Therefore, option (4) is correct.
### Pattern Recognition
The binding energy curve is a classic plateau curve peaking at Iron-56. Statements claiming monotonic trends across all masses in nuclear physics are generally false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
