An `alpha`-particle of `5MeV` energy strikes with a nucleus of uranium at stationary at an scattering angle of `108^(@)`. The nearest distance up to which `alpha`particle reaches the nuvles will be of the order of

Options

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

Answer

C

Explanation

At closest distance of apporach
Kinetic energy = Potential energy
`rArr 5 xx 10^(6) xx 1.6 xx 10^(-19) = (1)/(4pi epsilon_(0)) xx ((ze)(2e))/(r )`
For uranium `z = 92`, so `r = 5.3 xx 10^(-12) cm`