Practicing Success
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 |
$\sqrt{8}$ $\frac{1}{\sqrt{8}}$ 1 2 |
$\sqrt{8}$ |
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. $∴\frac{λ_p}{λ_α}=\frac{\sqrt{m_αq_α}}{m_pq_p}=\sqrt{\frac{4m}{m}×\frac{2e}{e}}=\sqrt{8}$ |