Define the distance of closest approach. An `alpha`-particle of kinetic enegy `'K'` is bombarded on a thin gold foil. The distance of the closest approach is `'r'` . What will be the distance of closest approach for an `alpha`-particle of double the kinetic energy ?

Options

A. `r_0/2`
B. `2r_0`
C. r
D. `r_0^2`

Answer

A

Explanation

The distance of the closest approach `r_0 prop 1/E` where E is the K.E. of the `alpha` particle. When E is doubled the distance of the closed approach will be `r_0/2 . [r=(2Ze^2)/(4piepsi_0E)]`