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?

Answer 1

Step:1 Find kinetic energy of the electron. Formula used: E = eV
Given,
voltage by which electron accelerate
$=50kV=50\times {10}^{3}V$
de Broglie wavelength $\lambda =?$
As we know charge on an electron $e=1.6\times {10}^{-19}C$
Mass of an electron, $m_e=9.11\times {10}^{-31}\text{kg}$
wavelength of yellow light, ${\lambda }_y=590\text{nm}=5.9\times {10}^{-7}m$
The kinetic energy of the electron is given by
K = eV
Substituting the values, we get
$K=1.6\times {10}^{-19}\times 50\times {10}^3$
$K=8\times {10}^{-15}$
Step:2 Find de Broglie wavelength of the electron.
Formula used:
$\lambda =\frac{h}{\sqrt{2m_{e}K}}$
De Broglie wavelength is given by the electron as: $\lambda =\frac{h}{\sqrt{2m_{e}K}}$
$\lambda =\frac{6.626\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 8\times {10}^{-15}}}$
$\lambda =5.467\times {10}^{-12}m$
This wavelength is ${10}^5$ times lesser than the wavelength of the yellow light. The resolving power (RP) of the microscope and the wavelength of the light used is inversely proportional.
Thus, RP of an electron microscope is about ${10}^5$ times that of an optical microscope. In practice, differences in other (geometrical) factors can change this comparison somewhat
Final Answer: $\lambda =5.467\times {10}^{-12}m$,
$(RP{)}_{\text{electron}}={10}^5(RP{)}_{\text{optical microscope}}$