Dual Nature of Matter

JEE Main level, numerical answer, medium

9 of 200

Question 9 of 200

The ratio of de Broglie wavelength of a deutron with kinetic energy EE to that of an alpha particle with kinetic energy 2E2 E, is n:1n: 1. The value of nn is ____\_\_\_\_.
(Assume mass of proton == mass of neutron) :

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Answer: 2

The de Broglie wavelength is λ=h2mE\lambda = \frac{h}{\sqrt{2mE}}.
For a deuteron (mass md=2mpm_d = 2m_p) with energy EE: λd=h2⋅2mp⋅E=h4mpE\lambda_d = \frac{h}{\sqrt{2 \cdot 2m_p \cdot E}} = \frac{h}{\sqrt{4m_p E}}
For an alpha particle (mass mα=4mpm_\alpha = 4m_p) with energy 2E2E: λα=h2⋅4mp⋅2E=h16mpE\lambda_\alpha = \frac{h}{\sqrt{2 \cdot 4m_p \cdot 2E}} = \frac{h}{\sqrt{16m_p E}}
The ratio: λdλα=16mpE4mpE=4=2\frac{\lambda_d}{\lambda_\alpha} = \frac{\sqrt{16m_p E}}{\sqrt{4m_p E}} = \sqrt{4} = 2

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