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JEE Main 2006 (AIEEE) · Atomic Models & Alpha-Particle Scattering

Atomic Models & Alpha-Particle Scattering question from JEE Main 2006 (AIEEE)

Q15JEE Main 2006 · AIEEEEasy

Atomic Models & Alpha-Particle Scattering

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$ . Then the distance of closest approach for the alpha nucleus will be proportional to

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Answer: (b)

$\frac{1}{2}mv^2 = \dfrac{k(2e)(Ze)}{r_0} \Rightarrow r_0 = \dfrac{4kZe^2}{mv^2}$ .
Thus $r_0 \propto \dfrac{1}{m}$ , $r_0 \propto \dfrac{1}{v^2}$ and $r_0 \propto Ze$ .
So the answer is (b) $\frac{1}{m}$ .

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