JEE Physics 2026 Question with Solution

Given: Energy of alpha particle = $7.7 \, \text{MeV}$, atomic number of gold = $79$, and $\frac{1}{4\pi\epsilon_0}=9\times10^{9}$ in SI units.

Find: The distance of closest approach $d$.

At the distance of closest approach, the entire initial kinetic energy of the alpha particle is converted into electrostatic potential energy.

By conservation of energy,

$$K = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(2e)}{d}$$

where $Z=79$.

Convert the given energy into joules:

$$K = 7.7 \, \text{MeV} = 7.7 \times 1.6 \times 10^{-13} \, \text{J}$$

Rearranging for $d$,

$$d = \frac{2Ze^2}{4\pi\epsilon_0 K} = \frac{2 Z e^2 (9 \times 10^9)}{K}$$

Substituting the values,

$$d = \frac{2 \times 79 \times (1.6 \times 10^{-19})^2 \times (9 \times 10^9)}{7.7 \times 1.6 \times 10^{-13}}$$ $$d = \frac{2 \times 79 \times 1.6 \times 9}{7.7} \times 10^{-16}$$ $$d \approx 295.48 \times 10^{-16} \, \text{m}$$ $$d \approx 2.95 \times 10^{-14} \, \text{m}$$

Therefore, the closest distance is $2.95 \times 10^{-14} \, \text{m}$. The correct option is C.

Given: An alpha particle of charge $2e$ approaches a gold nucleus of charge $Ze$ with $Z=79$.

Find: The nearest distance from the nucleus.

The alpha particle slows down because of Coulomb repulsion. At the turning point, its kinetic energy becomes zero, so its initial kinetic energy equals the electric potential energy.

Use

$$K = U$$ $$K = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(2e)}{d}$$

Now convert the energy scale:

$$1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}$$

so

$$7.7 \, \text{MeV} = 7.7 \times 1.6 \times 10^{-13} \, \text{J}$$

Hence,

$$d = \frac{(9 \times 10^9)(79e)(2e)}{7.7 \times 1.6 \times 10^{-13}}$$

Taking $e = 1.6 \times 10^{-19} \, \text{C}$,

$$d = \frac{2 \times 79 \times (1.6 \times 10^{-19})^2 \times 9 \times 10^9}{7.7 \times 1.6 \times 10^{-13}}$$

This gives

$$d \approx 2.95 \times 10^{-14} \, \text{m}$$

Thus, the alpha particle can come closest up to $2.95 \times 10^{-14} \, \text{m}$.