JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 10)
Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of $$\lambda_0$$. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :
3 $$\lambda_0$$
9 $$\lambda_0$$
$$\frac{\lambda_0}{\sqrt2}$$
$$\frac{\lambda_0}{2}$$
Explanation
When electron is accelerated through potential difference $V$, then
$$ \begin{aligned} & \text { K.E. }=\mathrm{eV} \\\\ & \Rightarrow \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\\\ & \therefore \lambda \alpha \frac{1}{\sqrt{\mathrm{V}}} \\\\ & \therefore \frac{\lambda}{\lambda_0}=\sqrt{\frac{20}{40}} \\\\ & \therefore \lambda=\frac{\lambda_0}{\sqrt{2}} \end{aligned} $$
$$ \begin{aligned} & \text { K.E. }=\mathrm{eV} \\\\ & \Rightarrow \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\\\ & \therefore \lambda \alpha \frac{1}{\sqrt{\mathrm{V}}} \\\\ & \therefore \frac{\lambda}{\lambda_0}=\sqrt{\frac{20}{40}} \\\\ & \therefore \lambda=\frac{\lambda_0}{\sqrt{2}} \end{aligned} $$
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