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  4. The de-Broglie wavelength of a particle of kinetic energy ' K ' is λ; the wavelength of the particle, if its kinetic energy is (K/4) is

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Q. The de-Broglie wavelength of a particle of kinetic energy ' $K$ ' is $\lambda$; the wavelength of the particle, if its kinetic energy is $\frac{K}{4}$ is

KCETKCET 2022Dual Nature of Radiation and Matter

Solution:

$\lambda \propto \frac{1}{\sqrt{ k }}$
$\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{ k _{2}}{ k _{1}}},=\sqrt{\frac{ k }{4 k }}=\frac{1}{2}$
$\lambda_{2}=2 \lambda$