An alpha nucleus of energy (1/2)mv2 bombards a heavy nuclear target of charge Ze . Then the distance of closest approach for the alpha nucleus will be proportional to

Question

An alpha nucleus of energy (1/2)mv2 bombards a heavy nuclear target of charge Ze . Then the distance of closest approach for the alpha nucleus will be proportional to (A) v2 (B) 1/m (C) 1/ Ze (D) 1 / v

Tardigrade Admin · Accepted Answer

At distance of closest approach relative velocity of two particles is

v

. Here target is considered as stationary, so

α

-particle comes to rest instantaneously at distance of closest approach. Let required distance is

r,

then from work energy-theorem.

0 - \frac{m v ^{2}}{2} = - \frac{1}{4 π ϵ _{0}} \frac{Z _{e} \times Z _{e}}{r}

r \propto \frac{1}{m}

\propto \frac{1}{v ^{2}}

\propto Z e^{2}

Q. An alpha nucleus of energy $\frac{1}{2} m v^{2}$ bombards a heavy nuclear target of charge $Z e$ . Then the distance of closest approach for the alpha nucleus will be proportional to

$v^{2}$

$1/ m$

1/ $Z e$

$1/ v$

Q. An alpha nucleus of energy 21​mv2 bombards a heavy nuclear target of charge Ze . Then the distance of closest approach for the alpha nucleus will be proportional to

v2

1/m

1/ Ze

1/v

Q. An alpha nucleus of energy $\frac{1}{2} m v^{2}$ bombards a heavy nuclear target of charge $Z e$ . Then the distance of closest approach for the alpha nucleus will be proportional to

$v^{2}$

$1/ m$

1/ $Z e$

$1/ v$