Question

A proton and an $\alpha -particle$ have the same de-Broglie wavelength. Determine the ratio of (i) their accelerating potentials (ii) their speeds

Answer 1

i) The relation between de Broglie wavelength $\lambda$ and accelerating potential $V$ is given by $\lambda =\frac{h}{\sqrt{2mqV}}$

So, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\left(\frac{m_{\alpha }}{m_p}\right)\left(\frac{q_{\alpha }}{q_p}\right)\left(\frac{V_{\alpha }}{V_p}\right)}$

Here, ${\lambda }_{\alpha }={\lambda }_p$, $\frac{m_{\alpha }}{m_p}=4/1$, $\frac{q_{\alpha }}{q_p}=2/1$

Thus, $1=\sqrt{\left(4\right)\left(2\right)\frac{V_{\alpha }}{V_p}}$

or $\frac{V_p}{V_{\alpha }}=\frac{8}{1}$

ii) Also , $\lambda =\frac{h}{p}=\frac{h}{mv}$

Thus, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{m_{\alpha }}{m_p}\frac{v_{\alpha }}{v_p}$

or $1=4\frac{v_{\alpha }}{v_p}$

or $\frac{v_p}{v_{\alpha }}=\frac{4}{1}$