The kinetic energy of an electron is tripled, then the de-Broglie wavelength associated with it, will change…

The kinetic energy of an electron is tripled, then the de-Broglie wavelength associated with it, will change by a factor
  1. $\frac{1}{3}$
  2. $3$
  3. $\sqrt{3}$
  4. $\frac{1}{\sqrt{3}}$

Solution

de-Broglie wavelength of an electron $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Or $\lambda \propto \frac{1}{\sqrt{\mathrm{K}}}$ where $\mathrm{K}$ is the kinetic energy of the electron and $\mathrm{m}$ is the mass. $\begin{aligned} & \therefore \frac{\lambda^{\prime}}{\lambda}=\frac{1}{\sqrt{3^{\mathrm{K}}}} \cdot \frac{\sqrt{\mathrm{K}}}{1}=\frac{1}{\sqrt{3}} \\ & \text { Or } \lambda^{\prime}=\frac{\lambda}{\sqrt{3}} \end{aligned}$ i.e., de-Broglie wavelength will decrease by a factor of $\frac{1}{\sqrt{3}}$. ~

Asked in: MHT CET 2022 (10 Aug Shift 1)

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