CBSE 2025, Set 3

Question

Define the 'distance of closest approach'. An $\alpha$-particle of kinetic energy $K$ is bombarded on a thin gold foil. Derive an expression for the 'distance of closest approach'.

Edvaya · Accepted Answer

The correct answer is Subjective. **Definition:** The distance of closest approach is the minimum distance between the $\alpha$-particle and the target nucleus at which the entire kinetic energy of the $\alpha$-particle gets converted into electrostatic potential energy. (Alternatively: It is the distance from the nucleus at which the alpha particle stops momentarily and then begins to retrace its path.)

**Derivation:**
Let the kinetic energy of the $\alpha$-particle be $K$.
The charge on the $\alpha$-particle is $q_1 = 2e$.
The charge on the target nucleus (gold) is $q_2 = Ze$.

At the distance of closest approach ($d$), the kinetic energy $K$ is entirely converted into potential energy $U$:

$$K = U$$

The electrostatic potential energy of the system is given by:
$$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d}$$

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

Rearranging for the distance of closest approach, $d$:
$$\boxed{d = \frac{2Ze^2}{4\pi\epsilon_0 K}}$$

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