Question

When an $\alpha$-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :

• A. $\frac{1}{\sqrt{m}}$
• B. $\frac{1}{m^2}$
• C. m
• D. $\frac{1}{m}$

Answer 1

Answer: (D)

$\frac{1}{m}$

Answer 2

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.
$\Rightarrow \, \frac{1}{2} mv^2 = \frac{KQq}{d} \, \Rightarrow \, d \propto \frac{1}{m}$