Question

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

Answer 1

De-Broglie wavelength of a particle=$\lambda =\frac{h}{p}$

But $p=\sqrt{2mK}$ where $K$ is the kinetic energy$=qV$

$⟹\lambda =\frac{h}{\sqrt{2mqV}}$

where $V$ is the accelerating potential of the particle.

i) for same de-Broglie wavelength,

$\frac{h}{\sqrt{2m_p{q}_p{V}_p}}=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }}{V}_{\alpha }}$

$⟹\frac{V_p}{V_{\alpha }}=\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}$

$=\frac{4}{1}\times \frac{2}{1}=8$

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

$⟹\frac{h}{m_p{v}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}$

$⟹\frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}=4$