**Question:**

An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

**Solution:**

Electrons are accelerated by a voltage, *V* = 50 kV = 50 × 103 V

Charge on an electron, *e* = 1.6 × 10−19 C

Mass of an electron, *m**e* = 9.11 × 10−31 kg

Wavelength of yellow light = 5.9 × 10−7 m

The kinetic energy of the electron is given as:

*E* = *eV*

= 1.6 × 10−19 × 50 × 103

= 8 × 10−15 J

De Broglie wavelength is given by the relation:

$\lambda=\frac{h}{\sqrt{2 m_{e} E}}$

$=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-15}}}$

$=5.467 \times 10^{-12} \mathrm{~m}$

This wavelength is nearly 105 times less than the wavelength of yellow light.

The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly 105 times that of an optical microscope.