Question

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:

• A. $\frac{\lambda_0}{2}$
• B. $3 \lambda_0$
• C. $\frac{\lambda_0}{\sqrt{2}}$
• D. $9 \lambda_0$

Answer 1

Answer: (C)

When electron is accelerated through potential difference $V$, then
K.E. $= eV $
$ \Rightarrow \lambda=\frac{ h }{\sqrt{2 m ( KE )}}=\frac{ h }{\sqrt{2 meV }} $
$ \therefore \lambda \alpha \frac{1}{\sqrt{ V }} $
$ \therefore \frac{\lambda}{\lambda_0}=\sqrt{\frac{20}{40}} $
$ \therefore \lambda=\frac{\lambda_0}{\sqrt{2}}$