An alpha nucleus of energy displaystylefrac-1-2-mv-2 bombards a heavy nuclear target of charge Ze- Then the distance of closest approach the alpha nucleus will be proportional todfrac-1-m-dfrac-1-v-4-dfrac-1-Ze-v-2

Question

Toppr Admin · Accepted Answer

Let the closest distance upto which the alpha particle approaches be $r$ .
At this distance the kinetic energy of the particle is zero.
Given: Initial kinetic energy of the particle, $K . E_{i} = \frac{1}{2} m v^{2}$
Also initial potential energy of the system is assumed to be zero.
Atomic number of alpha particle, $Z_{α} = 2$
Final potential energy of the system at closest distance, $P . E_{f} = \frac{K (Z e) 2 e}{r}$

Using conservation of energy : $K . E_{i} + P . E_{i} = K . E_{f} + P . E_{f}$
$∴$ $\frac{1}{2} m v^{2} + 0 = 0 + \frac{K 2 Z e^{2}}{r}$

$⟹$ $r = \frac{K 4 Z e^{2}}{m v^{2}} \propto \frac{1}{m}$ (where, $K = \frac{1}{4 π ϵ_{o}}$ )

An alpha nucleus of energy 12mv2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to1m1v41Zev2

An alpha nucleus of energy $\frac{1}{2} m v^{2}$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
$\frac{1}{m}$
$\frac{1}{v^{4}}$
$\frac{1}{Z e}$
$v^{2}$