The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \text{ Å}$ is (in $\text{ V}$)

The kinetic energy of an electron having de-Broglie wavelength $\lambda$ is ($h=$ Planck's constant,$m=$ mass of electron).

If the de Broglie wavelength of an electron is $2 \ nm$,then its kinetic energy is nearly (Planck's constant $= 6.6 \times 10^{-34} \ J \ s$ and mass of electron $= 9 \times 10^{-31} \ kg$) (in $eV$)

$A$ beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 \, Å$. The first maximum intensity occurs at $\theta = 60^{\circ}$. Then $E$ (in $eV$) is: (Planck constant $h = 6.64 \times 10^{-34} \, Js$, $1 \, eV = 1.6 \times 10^{-19} \, J$, electron mass $m = 9.1 \times 10^{-31} \, kg$)

An electron and a photon each have a wavelength of $1.00 \; nm$. Find
$(a)$ their momenta,
$(b)$ the energy of the photon,and
$(c)$ the kinetic energy of the electron.

If the de Broglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14} \, Hz,$ then the speed of the electron is equal to: (Speed of light $= 3 \times 10^8 \, m/s;$ Planck's constant $= 6.63 \times 10^{-34} \, J \cdot s;$ Mass of electron $= 9.1 \times 10^{-31} \, kg$)