When an electron is accelerated through a potential ' $\mathrm{V}$ ', the de-Broglie wavelength associated…

When an electron is accelerated through a potential ' $\mathrm{V}$ ', the de-Broglie wavelength associated with it is ' $\lambda$ '. When the accelerating potential is increased to $4 \mathrm{~V}$, its wavelength will be
  1. $\frac{\lambda}{4}$
  2. $\frac{\lambda}{2}$
  3. $\lambda$
  4. $2 \lambda$

Solution

$\begin{aligned} & \text { We know } \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\ & \Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{v}}}\end{aligned}$ $\begin{array}{ll} & \text { Given } \mathrm{V}=4 \mathrm{~V} \\ \therefore \quad & \lambda \propto \frac{1}{\sqrt{4 \mathrm{~V}}}=\frac{4}{2 \sqrt{\mathrm{V}}} \\ \therefore \quad & \lambda \text { reduces to } \frac{\lambda}{2}\end{array}$

Asked in: MHT CET 2023 (12 May Shift 2)

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