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

Answer

d

Explanation

If `d` is the distance of closet approach given, then the angular momentum `=mvd=10^(-33) J s`
`E=(1)/(2) mv^(2) =1 MeV =1.6 xx 10^(-33) J`
Momentum,
`P=sqrt(2m_nE)=sqrt(2 xx 1.6xx10^(27) xx1.6 xx 10^(-13))`
`=1.6 sqrt2 xx 10 ^(20) kg m s^(1)`
Distance of clossest approach,
`d=(10^(33))/(1.6sqrt2xx10^(-20))`
`=(1)/(1.6sqrt2) xx 10^(-13)=(100)/(1.6sqrt2)fm =0.44 fm`