1. 1/m 2. 1/√(m) 3. 1/m² 4. m

When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :

1. \(\frac{1}{m}\)

2. \(\frac{1}{\sqrt{m}}\)

3. \(\frac{1}{m^2}\)

4. \(m\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

Best answer

Answer: 1. \(\frac{1}{m}\)

Key idea: at the closest approach all kinetic energy becomes electrostatic potential energy.

\(\tfrac12mv^2=\dfrac{k(2e)(Ze)}{r_0}\Rightarrow r_0=\dfrac{4kZe^2}{mv^2}\propto\dfrac1m\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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