An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order of

Options

A. `1Å`
B. `10^(-10)cm`
C. `10^(-12) cm`
D. `10^(-15) cm`

Answer

C

Explanation

At the closest point of approach Initial `KE = ` Final PE
`:. 5 xx 10^(6) xx 1.6 xx 10^(-19) = (kq_(1)q_(2))/(r_(0))`
`r_(0) = (9 xx 10^(9) xx 92 xx 1.6 xx 10^(-19) xx 2 xx 1.6 xx 10^(19))/(5 xx 10^(6) xx 1.6 xx 10^(-19))`
`= 5.3 xx 10^(-14) m = 5.3 xx 10^(-12) cm`