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 :
Solution
When electron is accelerated through potential difference $V$, then<br/><br/>
$$
\begin{aligned}
& \text { K.E. }=\mathrm{eV} \\\\
& \Rightarrow \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\\\
& \therefore \lambda \alpha \frac{1}{\sqrt{\mathrm{V}}} \\\\
& \therefore \frac{\lambda}{\lambda_0}=\sqrt{\frac{20}{40}} \\\\
& \therefore \lambda=\frac{\lambda_0}{\sqrt{2}}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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