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?

(N/A) Electrons are accelerated by a voltage,$V = 50\; kV = 50 \times 10^{3}\; V$.
Charge on an electron,$e = 1.6 \times 10^{-19}\; C$.
Mass of an electron,$m_{e} = 9.11 \times 10^{-31}\; kg$.
Wavelength of yellow light $\lambda_{yellow} \approx 5.9 \times 10^{-7}\; m$.
The kinetic energy of the electron is $E = eV = 1.6 \times 10^{-19} \times 50 \times 10^{3} = 8 \times 10^{-15}\; J$.
The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m_{e} E}}$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-15}}} \approx 5.467 \times 10^{-12}\; m$.
This wavelength is nearly $10^{5}$ times smaller than the wavelength of yellow light.
Since the resolving power of a microscope is inversely proportional to the wavelength,the resolving power of an electron microscope is approximately $10^{5}$ times greater than that of an optical microscope.