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 $\frac{d}{2}$ towards a stationary nucleus of double the charge, the velocity of projection of the $\alpha$- particle has to be

• A. $\sqrt{2}v$
• B. $2v$
• C. $4v$
• D. $\sqrt{6}v$

Answer 1

The correct option is B $2v$

From work-energy theorem,

$(KE{)}_{\alpha }$ = P.E of Nucleus and $\alpha -$ particle

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_n{q}_{\alpha }}{d}\right)........\left(1\right)$

For, $q_n^{{}^{\prime }}=2q_n{d}^{{}^{\prime }}=\frac{d}{2}$

$\frac{1}{2}mv{{}^{\prime }}^2=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_n^{{}^{\prime }}{q}_{\alpha }}{d^{{}^{\prime }}}\right)........\left(2\right)$

On dividing (2) $÷$ (1)

${\left(\frac{v^{\prime }}{v}\right)}^2=\frac{q_{n^{\prime }}}{q_n}\times \frac{d}{d^{{}^{\prime }}}$

${\left(\frac{v^{\prime }}{v}\right)}^2=\frac{2q_n}{q_n}\times \frac{d}{\left(\frac{d}{2}\right)}=4$

$\therefore v^{{}^{\prime }}=2v$