Question

A proton and a ${\text{Li}}^{+}$ ion are moving towards a gold foil. If their distance of the closest approach are equal, find $\frac{v_{L{I}^{+}}}{v_{\alpha }}$.

• A. $\frac{2}{\sqrt{3}}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B $\sqrt{\frac{2}{3}}$

The distance of closest approach is given by,

$d=\frac{4KZ{e}^2}{m{v}^2}$

As, $d_{L{I}^{+}}=d_{\alpha }$

$\Rightarrow \frac{4KZ{e}^2}{m_{L{I}^{+}}{v}_{L{I}^{+}}^2}=\frac{4KZ{e}^2}{m_{\alpha }{v}_{\alpha }^2}$

On simplifying, we get,

$\frac{v_{L{I}^{+}}}{v_{\alpha }}=\sqrt{\frac{m_{\alpha }}{m_{L{I}^{+}}}}=\sqrt{\frac{4m_p}{6m_p}}=\sqrt{\frac{2}{3}}$

Hence, $\left(B\right)$ is the correct answer.