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

Answer

A

Explanation

For the distance (r ) of the closest approach , the K.E. is converted into P.E.
Initial K.E. = P.E.
`therefore 5MeV = (1)/(4piepsi_0)xx(q_1q_2)/(r^2) xxr = (1)/(4piepsi_0) (q_1q_2)/(r )`
`therefore q_1 = 92e` (Uranium nucleus ) and `q_2=2e(alpha "particle")`
`5xx10^6 xxe = ((9xx10^9)xx(92e) xx(2e))/(5xx10^6)`
`=(9xx10^3xx92xx2xx1.6xx10^(-19))/(5)`
`r=5.3xx10^(-14) m = 5.3xx10^(-12) cm`
`therefore` r is of the order of `10^(-12)` cm