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Physics · JEE | Nuclei
हिन्दीIf an alpha particle of kinetic energy \( E \) approaches a gold nucleus, the distance of closest approach \( d \) is proportional to:
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Correct answer · Option B
\( \frac{1}{E} \)
Explanation
By energy conservation, kinetic energy equals electrostatic potential energy at closest approach: \( E = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{d} \), hence \( d \propto \frac{1}{E} \).
