Question

If a proton and α -particle are accelerated through the same potential difference, the ratio of de-Broglie wavelengths $\lambda_p$ and $\lambda_{\alpha}$ is
(a) 3
(b) $2\sqrt{2}$
(c) 1
(d) 2

Answer 1

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Answer 2

Δ=h/√2mK or it can be written as Δ=h√2mqV bcoz as we know potential difference I.e . V =K. E/q now for same potential difference de broglie wavelength I. e .  Δ is proportional to    1/√mq so Δp/Δα =√4m 2e/√2me =2