The distance of closest approach of an alpha-particle fired towards a nucleus with momentum p is r. What will be the distance of closest approach when the momentum of alpha-particle is `2p`?

Options

A. r/2
B. 2r
C. 4r
D. r/4

Answer

D

Explanation

At the distance of closed approach, r
`" " K=(1)/(4piepsi_(0))((2e)(Z2))/(r)rArrr=(2Ze^(2))/(4piepsi_(0)K)`
where Ze= charge of the nucleus
`" " 2e` = change of the alpha particle
`" "K` = kinetic energy of the alpha
`:. " " K=(p^(2))/(2m)`
where , p is the momentum of the `alpha` and m is mass of the electron
`:. " " r=(2Ze^(2)2m)/(4piepsi_(0)p^(2))orrprop(1)/(p^(2))`
`" " (r')/(r)=((p)/(p'))^(2)=((p)/(2p))^(2)=(1)/(4)rArrr'=(r)/(4)`