Question

Electron beam used in an electron microscope, when accelerated by a voltage of $20kV$, has a de-Broglie wavelength of ${\lambda }_0$. If the voltage is increased to $40kV$, 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. $=eV$
$\Rightarrow \lambda =\frac{h}{\sqrt{2m(KE)}}=\frac{h}{\sqrt{2meV}}$
$\therefore \lambda \alpha \frac{1}{\sqrt{V}}$
$\therefore \frac{\lambda }{{\lambda }_0}=\sqrt{\frac{20}{40}}$
$\therefore \lambda =\frac{{\lambda }_0}{\sqrt{2}}$