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MHT CET 2026 (20 Apr, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from MHT CET 2026 (20 Apr, Shift 2)

Q4MHT CET 2026 · 20 APR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

The ratio of de-Broglie wavelength of an $\alpha$ -particle and proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$ . The value of $m$ is

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Answer: (b)

$\lambda = \dfrac{h}{\sqrt{2mqV}}$ , so for the same $V$ , $\lambda \propto \dfrac{1}{\sqrt{mq}}$ .
$\dfrac{\lambda_\alpha}{\lambda_p} = \sqrt{\dfrac{m_p e}{4m_p\cdot 2e}} = \dfrac{1}{\sqrt{8}}$ .
So $m = 8$ ; the answer is (b) 8.

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