An electron microscope uses electrons accelerated by a voltage of 50 \mathrm{kV}. Determine the de B roglie 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?
An electron microscope uses electrons accelerated by a voltage of 50 \mathrm{kV}. Determine the de B roglie 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?

Electrons are accelerated by a voltage of 50 \mathrm{kV}

Charge on an electron, e=1.6 \times 10^{-19} \mathrm{C}

Mass of the electron, m_{e}=9.11 \times 10^{-31} \mathrm{~kg}

Wavelength of the yellow light is given as 5.9 \times 10^{-7} \mathrm{~m}

The kinetic energy of the electron, E=e V

=\left(1.6 \times 10^{-19}\right) \times\left(50 \times 10^{3}\right)

=8 \times 10^{-15} \mathrm{~J}

De Broglie wavelength of electron is given as

\lambda=\frac{h}{\sqrt{2 m_{c} E}}

\lambda=\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}

The wavelength of the blue light is 10^{5} times shorter than that of the yellow light. The wavelength of the light used and the resolving power of the microscope are inversely related. As a result, the electron microscope’s resolving power is 10^{5} times larger than that of an optical microscope.