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

If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes
  1. $2$ times
  2. $\frac{1}{\sqrt{2}}$ times
  3. $4$ times
  4. $\sqrt 2$ times

Solution

$\begin{aligned} & \text { } \lambda \propto \frac{1}{\sqrt{\mathrm{K} . E}} \Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{KE})_2}{(\mathrm{KE})_1}} \\ & \mathrm{KE}_2=\frac{(\mathrm{KE})_1}{2} \\ & \text { Thus, } \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2}} \Rightarrow \lambda_2=\sqrt{2} \lambda_1\end{aligned}$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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