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. `10^(-12) cm`
B. `10^(-10)cm`
C. `10^(-20) cm`
D. `10^(-15)cm`

Answer

A

Explanation

Distance of closest approach `r_0 = (Ze (2e ))/( 4 pi epsi_0 (1/2 mv^2))`
energy ` E= 5 xx 10^6 xx 1.6 xx 10^(-19) J`
` therefore r_ 0 = ( 9 xx 10^( 9) xx ( 92 xx 1.6 xx 10^(-19) ) (2 xx 1.6 xx 10^(-19) )/( 5 xx 10^6 xx 1.6 xx 10^(-19))`
`implies r= 5.2 xx 10^(-14) m= 5.3 xx 10^(12) cm`