In an electron microscope,which nature of the electron is used?

An electron accelerated by a potential difference $V$ has a de-Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $9V$,its de-Broglie wavelength will be

$A$ proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ are the kinetic energies of the proton and electron respectively,then choose the correct relation:

The de-Broglie wavelength of an electron having $80 eV$ energy is nearly ($1 eV = 1.6 \times 10^{-19} J$,Mass of the electron $= 9 \times 10^{-31} kg$,Planck's constant $= 6.6 \times 10^{-34} J-s$). (in $Å$)

What is the de-Broglie wavelength of a bullet of mass $0.033$ kg travelling at the speed of $1$ km/s? $(h = 6.6 \times 10^{-34} \text{ Js})$

An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few $eV$. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ($c =$ speed of light in vacuum).