If the kinetic energy of a particle in motion is decreased by $36 \%$, the increase in de Broglie wavelength of the particle is

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Aicharya · Accepted Answer

$\mathrm{K}_2=(1-0.36) \mathrm{K}_1=0.64 \mathrm{~K}_1$
De-Broglie wavelength of the particle is $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}} \Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{k}}} \\ & \therefore \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}=\sqrt{\frac{\mathrm{k}_1}{0.64 \mathrm{k}_1}}=1.25 \\ & \therefore \lambda_2=1.25 \lambda_1 \end{aligned}$ $\therefore \quad \%$ increase in de-Broglie wavelength is $\% \Delta \lambda=\frac{\Delta \lambda}{\lambda_1} \times 100=\frac{1.25 \lambda_1-\lambda_1}{\lambda_1} \times 100=25 \%$

If the kinetic energy of a particle in motion is decreased by $36 \%$, the increase in de Broglie wavelength…

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