Question

An alpha nucleus of energy $\frac{1}{2}$ m ${\vartheta }^2$ bombards a heavy nuclear of charge Ze. Then the distance of closset approach for the alpha nucleus will the proportional to

• A. $\frac{1}{\vartheta }$
  
• B. $\frac{1}{Ze}$
  
• C. $\vartheta$
  
• D. $\frac{1}{m}$

Answer 1

The correct option is D $\frac{1}{m}$

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_0}\frac{\left(Ze\right)\left(2e\right)}{d}$
$d\propto \frac{1}{m}$ ; $d\propto \frac{1}{v^2}$