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?

Hint: From the acceleration voltage get the kinetic energy of the accelerated electron. Using this to get the momentum of the electron and de-Broglie equation will give you the wavelength of the electrons. To compare the resolving power always remember that it is inversely proportional to the wavelength of light used.

Formula Used:
Kinetic energy of accelerated electron is given by:
$E = e.V$ -----(1)
Where,
E denotes the kinetic energy of the electron,
e is the charge of an electron,$e = 1.6 \times {10^{ - 19}}C$.
V is the accelerating potential applied.

Momentum of accelerated electron is given by:
$p = \sqrt {2mE} $ -----(2)
Where,
p is the momentum of the electron,
m is mass of an electron, $m = 9.1 \times {10^{ - 31}}kg$.

De-Broglie equation:
$\lambda = \dfrac{h}{p}$ ----(3)
Where,
$\lambda $is the wavelength associated to the particle,
h is the planck's constant, $h = 6.63 \times {10^{ - 34}}J.s$.

Complete step by step answer:
Given: Acceleration voltage of the electron is 50kV i.e. $V = 50kV = 5 \times {10^4}V$.
To find: Wavelength associated with the electron.
Comparison of resolution power with a microscope using yellow light.

> Step 1
First, using eq.(1) and eq.(2) in eq.(3) get an expression for $\lambda $as:
\[\lambda = \dfrac{h}{p} = \dfrac{h}{{\sqrt {2mE} }} = \dfrac{h}{{\sqrt {2mEV} }}\] ---(4)

> Step 2
Now, substitute the value of V to get $\lambda $:
$
  \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}} }}m \\
   = 5.467 \times {10^{ - 12}}m \\
 $

> Step 3
Hence, you got the wavelength associated with the electron as $5.467 \times {10^{ - 12}}m$. Wavelength of yellow light is ${\lambda _y} = 590nm = 5.9 \times {10^{ - 7}}m$. Now, the resolving power of a microscope is inversely proportional to the wavelength of light used.

Hence, the ratio of resolving power for the electron and optical microscope will be: $\dfrac{{{\lambda _y}}}{\lambda } = \dfrac{{5.9 \times {{10}^{ - 7}}m}}{{5.467 \times {{10}^{ - 12}}m}} \simeq {10^5}$.


Hence, De-Broglie wavelength associated with the electrons is $5.467 \times {10^{ - 12}}m$. The resolving power of an electron microscope will be ${10^5}$times of that of an optical microscope that uses yellow light.

Note: As it’s evident you have noticed that electron microscopes have significantly larger resolving power, very tiny biological specimens like tissues, cells or even some molecules are observed using electron microscopes.