Question

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

• A. $v^2$
• B. $\frac{1}{m}$
• C. $\frac{1}{v^4}$
• D. $\frac{1}{Ze}$

Answer 1

Answer: (B)

$\frac{1}{m}$

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, $Z_{\alpha }=2$

Final potential energy of the system at closest distance, $P.E_f=\frac{K\left(Ze\right)2e}{r}$

Using conservation of energy : $K.E_i+P.E_i=K.E_f+P.E_f$

$\therefore$ $\frac{1}{2}m{v}^2+0=0+\frac{K2Z{e}^2}{r}$

$⟹$ $r=\frac{K4Z{e}^2}{m{v}^2}\propto \frac{1}{m}$ (where, $K=\frac{1}{4\pi {ϵ}_o}$)