The closet distance of approach of an `alpha`-particle travelling with a velocity v to a certain nucleus is x.the distance of closet approach of `alpha`-particle travelling with a velocity 3 v to the same nucleus is

Options

A. `X/5`
B. `X/6`
C. `X/9`
D. `X/10`

Answer

c

Explanation

The distance of closests approach of `alpha ` particle is ,
`r_(0)=(1)/(4piepsilon_(0))(2Ze^(2))/((1)/(2)mv^(2))`
`r_(0)prop(1)/v^(2)`
`r_(1)/(r_2)(v_(2)/v_(1))^(2)`
`(x)/r_(2)=((3v)/(v)^(2))`
`thereforer_(2)=(x)/(9)`