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(D) The de-Broglie wavelength $\lambda$ is related to kinetic energy $K.E.$ by the formula $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.This implies $\lambda

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$\sqrt{2}$ times

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(D) The de-Broglie wavelength $\lambda$ is related to kinetic energy $K.E.$ by the formula $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.This implies $\lambda \propto \frac{1}{\sqrt{K.E.}}$Given that the kinetic energy is reduced to half,$(K.E.)_2 = \frac{(K.E.)_1}{2}$.Using the ratio $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{(K.E.)_1}{(K.E.)_2}} = \sqrt{\frac{(K.E.)_1}{(K.E.)_1 / 2}} = \sqrt{2}$.Therefore,$\lambda_2 = \sqrt{2} \lambda_1$.

If the kinetic energy of a particle is reduced to half,the de-Broglie wavelength becomes

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