Question

The closest distance of approach of an $\alpha$ - particle travelling with a velocity V towards a stationary nucleus is 'd'. For the closest distance to become $\frac{d}{3}$ towards a stationary nucleus of double the charge, the velocity of projection of the $\alpha$ - particle has to be

Answer 1

$\frac{1}{2}m{v}^2=\frac{K.2l.2l}{d}\Rightarrow d=\frac{K.2\tau e^2}{\frac{1}{2}m{v}^2}Now,\frac{d}{3}=\frac{K.2l.2\tau l}{\frac{1}{2}m{v}^2}\Rightarrow \frac{K2\tau e^2}{3.\frac{1}{2}m{v}^2}=\frac{K.2e.2ze}{\frac{1}{2}m{v}^{{}^{\prime }2}}\Rightarrow \frac{1}{3V^2}=\frac{2}{V^{{}^{\prime }2}}\Rightarrow V^{{}^{\prime }2}=6V^2\therefore V^{{}^{\prime }}=\sqrt{6}V$