A proton and an alpha, particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is
Answer & explanation
Correct answer: option 1
When a charged particle of charge q and mass m is accelerated under a potential difference V, let v be velocity acquired by the particle. Then
$qV=\frac{1}{2}mv^2$ or $mv=\sqrt{2mqV}$
de Broglie wavelength, $λ=\frac{h}{mv}=\frac{h}{\sqrt{2mqV}}$
$λ∝\frac{1}{\sqrt{mq}}$ for the same value of V.
$∴\frac{λ_p}{λ_α}=\frac{\sqrt{m_αq_α}}{m_pq_p}=\sqrt{\frac{4m}{m}×\frac{2e}{e}}=\sqrt{8}$