An electron microscope uses which property of the electron?

$A$ nucleus $A$,with a finite de Broglie wavelength $\lambda_A$,undergoes spontaneous fission into two nuclei $B$ and $C$ of equal mass. $B$ flies in the same direction as that of $A$,while $C$ flies in the opposite direction with a velocity equal to half of that of $B$. The de Broglie wavelengths $\lambda_B$ and $\lambda_C$ of $B$ and $C$ are respectively

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?

An electron and a proton are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{p}$ to $\lambda_{e}$ is $[m_{e} = \text{mass of electron}, m_{p} = \text{mass of proton}]$

An electron is accelerated through a potential difference of $10,000 \ V$. Its de-Broglie wavelength is nearly: $(m_{e} = 9 \times 10^{-31} \ kg)$

Both an electron and a photon have the same de-Broglie wavelength of $1.2 \ \text{Å}$. The ratio of their energies is nearly