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 the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? $\left( {{\lambda }_{y}}=5.9\times {{10}^{-7}}m \right)$

Answer 1

Find the momentum (p=mv) of a particle in terms of kinetic energy ($K=\dfrac{1}{2}m{{v}^{2}}$). The de-Broglie wavelength of a particle is given as $\lambda =\dfrac{h}{p}$. Use this formula to find the wavelength of the electron. Also you use the work energy theorem to find the relation between kinetic energy and potential difference. The resolving power of a microscope is inversely proportional to wavelength.

Formula used:
$\lambda =\dfrac{h}{p}$
$K=\dfrac{1}{2}m{{v}^{2}}$
W=K
p=mv

Complete step-by-step answer:
According to de-Broglie every particle or body has a wave nature. The wavelength of the wave is called de-Broglie wavelength($\lambda$). The value of the wavelength is inversely proportional to the momentum of the particle and it is given as $\lambda =\dfrac{h}{p}$ ….
(i).
Here, p is the momentum of the particle and h is Planck’s constant.
In this case, the particle is an electron. It is given as acceleration by a voltage of 50kV, which is equal to $5\times {{10}^{4}}V$.
The voltage or potential difference will do a work which will change the kinetic energy of the electron. According to the work energy theorem, the work done is equal to the change in the kinetic energy of the particle.
i.e. W=K.
When a charge q passes through a potential difference of V. The work done on the charge is equal to W=qV.
Therefore, we get W=qV=K
We know that $K=\dfrac{1}{2}m{{v}^{2}}$
Multiple m on both the sides.
$mK=\dfrac{1}{2}{{m}^{2}}{{v}^{2}}$
and p=mv.
Therefore, $mK=\dfrac{1}{2}{{p}^{2}}$
$\Rightarrow p=\sqrt{2mK}$
And we found that K=qV.
$\Rightarrow p=\sqrt{2mqV}$.
Substitute the value of p in equation (i).
$\Rightarrow \lambda =\dfrac{h}{\sqrt{2mqV}}$ ….. (ii).
The value of Planck’s constant $h=6.63\times {{10}^{-34}}Js$,
Mass of an electron $m=9.1\times {{10}^{-31}}kg$
The magnitude of charge on an electron $e=1.6\times {{10}^{-19}}C$
The potential difference $V=5\times {{10}^{4}}V$
Substitute all the values in equation (ii).
$\Rightarrow \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 1.6\times {{10}^{-19}}\times 5\times {{10}^{4}}V}}$
$\Rightarrow \lambda =\dfrac{6.63\times {{10}^{-34}}}{120.66\times {{10}^{-24}}}=5.46\times {{10}^{-12}}m$
Therefore, the de-Broglie wavelength of the electrons is $5.46\times {{10}^{-12}}m$ .
It is given that the optical microscope uses yellow light whose wavelength is ${{\lambda }_{y}}=5.9\times {{10}^{-7}}m$.
This means that the wavelength of light is almost ${{10}^{5}}$ times larger than the de-Broglie wavelength of the electrons.
The resolving power is inversely proportional to the wavelength of the wave. Therefore, when we compare the resolving power of the electron microscope and the optical microscope, we get that the resolving power of an electron microscope is ${{10}^{5}}$ greater than the resolving power of the optical microscope.

Note: According to de-Broglie, all objects in the universe have a wave nature. However, we do not see the objects moving like a wave.
Large objects that we see daily have a wave nature but it cannot be seen by our naked eyes. This is because the wavelengths of these waves are very small compared to its size.
If we see equation (ii), the wavelength is inversely proportional to the mass of the object and the value of h is very small. Therefore, the wavelength will be very small.