If an α - particle of mass m, charge q and velocity v is incident on a nucleus of charge Q and mass m, then the distance of closest approach is

Question

If an α - particle of mass m, charge q and velocity v is incident on a nucleus of charge Q and mass m, then the distance of closest approach is (A) ( Qq /4 π ε0 m 2) (B) ( Qq /2 π ε0 mv 2) (C) ( Qq mv 2/2) (D) ( Qq / mv 2)

Tardigrade Admin · Accepted Answer

For the distance of closest approach kinetic energy will totally be converted to potential energy. Hence, (1/2) mv 2=(1/4 π ε0) ( Qq / r 0) ⇒ r 0=( Qq /2 π ε0 mv 2)

Q. If an $α$ - particle of mass $m$ , charge $q$ and velocity $v$ is incident on a nucleus of charge $Q$ and mass $m$ , then the distance of closest approach is

$\frac{Qq}{4 π ε _{0} m ^{2}}$

$\frac{Qq}{2 π ε _{0} m v ^{2}}$

$\frac{Qq m v ^{2}}{2}$

$\frac{Qq}{m v ^{2}}$

For the distance of closest approach kinetic energy will totally be converted to potential energy.
Hence, $\frac{1}{2} m v^{2} = \frac{1}{4 π ε _{0}} \frac{Qq}{r _{0}}$
$\Rightarrow r_{0} = \frac{Qq}{2 π ε _{0} m v ^{2}}$

Q. If an α - particle of mass m, charge q and velocity v is incident on a nucleus of charge Q and mass m, then the distance of closest approach is

4πε0​m2Qq​

2πε0​mv2Qq​

2Qqmv2​

mv2Qq​