question_answer

An\[\alpha -\]particle of energy\[5\text{ }MeV\]is scattered through\[180{}^\circ \,\]by a fixed uranium nucleus. The distance of the closest approach is of the order of                         [AIEEE 2004]

A) \[1{\AA}\]

B) \[{{10}^{-10}}cm\]

C) \[{{10}^{-12}}cm\]

D) \[{{10}^{-15}}cm\]

Correct Answer: C

Solution :

[c] According to law of conservation of energy, kinetic energy ofparticle

= potential energy of a-particle at distance of closest approach

i.e.,