The distance of the closest approach of an `alpha`- particle fired at a nucleus with kinetic energy K is `r_(0)`. The distance of the closest approach when the `alpha`-particle is fire at the same nucleus with kinetic energy 2K will be

Options

A. `(r_(0))/(2)`
B. `(r_(0))/(4)`
C. `4r_(0)`
D. `2r_(0)`

Answer

A

Explanation

Distance of closest approach
`r_(0)=(2Ze^(2))/(4pi epsi_(0)xxK)` where `2kze^(2)` are constant
`:.r_(0)= prop (1)/(K)`
`:. ((r_(0))_(2))/((r_(0))_(1))=(K_(1))/(K_(2))=(K)/(2K)`
`:.(r_(0))_(2)=((r_(0))_(1))/(2)=(r_(0))/(2) [ :. (r_(0))_(1)=r_(0)]`