A neutron of energy 1 MeV and mass `1.6 xx 10^(-27)` kg passes a proton at such a distance that the angular momentum of the neutron relative to the proton approximately equals `10^(-33) J s`. The distance of closest approach neglecting the interaction between particles si

Options

A. `0.44` nm
B. 0.44 mm
C. 0.44 Å
D. 0.44 fm

Answer

D

Explanation

If d is the distance of closedt approach given , then the angular mometum = mvd = `10^(-33) Js`
`E = (1)/(2) mv^(2) = 1 Me V = 1.6 xx 10^-13` J
Momentum ,
`P = sqrt(2 m_(n) E) = sqrt(2 xx 1.6 xx 10^(-27) xx 1.6 xx 10^(-13)) = 1.6 sqrt2 xx 10^(-20) kg m s^(-1)`
Distance of closest approach ,
`d = (10^(-33))/(1.6 sqrt2 xx 10^(-20))`
`(1)/(1.6 sqrt2) xx 10^(-13) = (100)/(1.6 sqrt2) fm = 0.44` fm