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AIPMT PRELIMS2010Physics-Nuclear Physics

AIPMT PRELIMS 2010 Physics Distance of Closest Approach MCQ Question

Type: MCQ-conceptual-Medium-Class 12

A 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

A

$\frac{1}{v^4}$

B

$\frac{1}{Ze}$

C

$v^2$

D

$\frac{1}{m}$

Correct Answer

Option D

Detailed Explanation

The distance of closest approach $r_0$ is given by $r_0 = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{\frac{1}{2} mv^2}$ . It is inversely proportional to the mass $m$ of the alpha particle.

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