An particle of energy 5 MeV is scattered through 180° by gold nucleus. The distance of closest approach is of the order of

Options

A. `10^(-12)cm`
B. `10^(-16)cm`
C. `10^(-10)cm`
D. `10^(-14)cm`

Answer

A

Explanation

Gain in K.E. = Potential energy
`(1)/(2) mv^(2) = (K(2e) (79e))/(r)`
`(1)/(2) mv^(2)= 5MeV= 5 xx 10^(6) eV`
`r= (K (2e) (79e))/(5 xx 10^(-6) xx 1.6 xx 10^(-19))= (9 xx 10^(9) xx 2 xx 1.6 xx 10^(-19) xx 79 xx 1.6 xx 10^(-19))/(5 xx 1.6 xx 10^(-13)) ~~ 10^(-12) cm`