What is the theoretical formula for the de-Broglie wavelength of a particle?

Find the de-Broglie wavelength of an electron with kinetic energy of $ 120 eV $. (in $pm$)

The wavelength of a charged particle of mass $8.0 \times 10^{-31} \ kg$, charge $1.6 \times 10^{-19} \ C$ and kinetic energy $3 \ keV$ will be (Planck constant, $h = 6.4 \times 10^{-34} \ Js$) (in $\text{Å}$)

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-rays is

Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will