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de Broglie wavelength of a neutron particle having kinetic energy E is $λ$ . What will be the de Broglie wavelength of a neutron particle having energy 9E?

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$\frac{λ}{6}$

$\frac{λ}{2}$

$\frac{2 λ}{3}$

$\frac{λ}{3}$

answer is D.

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Detailed Solution

$\frac{p^{2}}{2 m} = E \Rightarrow P = \sqrt{2 m E} \Rightarrow λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}}$

$∴ \frac{λ^{1}}{λ} = \frac{\sqrt{2 m E}}{\sqrt{2 m E^{1}}} = \sqrt{\frac{E}{9 E}} = \frac{1}{3} \Rightarrow λ^{1} = λ / 3$

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