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JEE Main 2021 (25 Feb, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2021 (25 Feb, Shift 1)

Q101JEE Main 2021 · 25 FEB · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda$ p respectively. The ratio $\frac{\lambda_p}{\lambda_\alpha}$ is:

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Answer: (b)

$\lambda = \dfrac{h}{\sqrt{2mqV}}$ , same $V$ for both.
$\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4 \times 2} = \sqrt{8} \approx 2.83$
So the answer is (b) 2.8.

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