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 fired at the same nucleus with kinetic energy 2K will be

Q. The distance of closest approach of an

α

-particle fired at nucleus with momentum

P

d

. The distance of closest approach when the

α

-particle is fired at same nucleus with momentum

3 P

will be:

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An α- particle accelerated through V volt is fired towards a nucleus. Its distance of closest approach is r. If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be:

The correct option is C rFor α - particle:12mv2=q1q24πϵ b2eV= 2e×q24πϵ0 rq2= 4πϵ0rV for protoneV= q1q24πϵ beV=e q24πϵ bq2=4πϵ b V∴b=r

An $α$ - particle accelerated through V volt is fired towards a nucleus. Its distance of closest approach is $r$ . If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be:

The correct option is C $r$
For $α$ - particle:
$\frac{1}{2} m v^{2} = \frac{q_{1} q_{2}}{4 π ϵ b}$
$2 e V = \frac{2 e \times q_{2}}{4 π ϵ_{0} r}$
$q_{2} = 4 π ϵ_{0} r V$ for proton
$e V = \frac{q_{1} q_{2}}{4 π ϵ b}$
$e V = \frac{e q_{2}}{4 π ϵ b}$
$q_{2} = 4 π ϵ b V$
$∴ b = r$