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 additional energy that should be supplied to a moving electron to reduce its de Broglie wavelength from $1 \,nm$ to $0.5 \,nm$?

The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?

If the mass of a neutron is $1.7 \times 10^{-27} \; kg$,then the de-Broglie wavelength of a neutron with energy $3 \; eV$ is (given $h = 6.6 \times 10^{-34} \; J \cdot s$):

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. If $\lambda_1$ and $\lambda_2$ are the de Broglie wavelengths of the proton and the photon respectively,then find the ratio $\lambda_1 / \lambda_2$.

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 $eV$)