Question

An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? $\left({\lambda }_y=5.9\times {10}^{-7}\text{m}\right)$

Answer 1

Given:- Accelerating voltage=$V=50kV$

$=50\times {10}^{3}V$

Solution:-Using de-broglie's equation.

$\lambda =\frac{h}{p}⟶\left(1\right)$

$\lambda$=Wavelength of moving particle.

$P$=Momentum of moving particle.

$h$=Planck's constant$=6.63\times {10}^{-3}Js$

now from mechanics we know,

$P=\sqrt{2mk}=\sqrt{2mqV}$

Where,$P$=momentum of particle

$M$=mass of particle.

$K$=Kinetic energy of particle.

$V$= Electric potential of particle( Accelerating voltage).

$q$=charge of particle.

In the case of electron,$q=e=1.6\times {10}^{-19}C$

Substitution of $\left(1\right)$

$\therefore \lambda =\frac{h}{\sqrt{2meV}}$

$\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 50\times {10}^3}}$

$\lambda =\frac{6.63\times {10}^{-34}}{120.66\times {10}^{-24}}$

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

Whereas wavelength of yellow light, ${\lambda }_y=590nm$

$=5.9\times {10}^{-7}m$

The wavelength of the accelerated electron is about ${10}^5$ times less than the yellow light.

Since resolving power of a microscope is inversely proportional to the wavelength of light hence the resolving power of the electron microscope is about ${10}^5$ times more than that of an optical microscope.