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JEE Main 2022 (27 Jun, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2022 (27 Jun, Shift 1)

Q83JEE Main 2022 · 27 JUN · SHIFT 1Easy

Matter Waves: de Broglie Wavelength

An $\alpha$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $(\lambda_\alpha : \lambda_{C12})$ is:

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Answer: (b)

Same kinetic energy: $\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{m}}$ .
$\dfrac{\lambda_\alpha}{\lambda_C} = \sqrt{\dfrac{m_C}{m_\alpha}} = \sqrt{\dfrac{12}{4}} = \sqrt{3}$
So the answer is (b) $\sqrt{3} : 1$ .

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