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?

$A$ proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is $2000$ times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(\lambda_{p})$ and that of the electron $(\lambda_{e})$?

The de-Broglie wavelength of an electron is $0.4 \times 10^{-10} \ m$ when its kinetic energy is $1.0 \ keV$. Its wavelength will be $1.0 \times 10^{-10} \ m$ when its kinetic energy is: (in $keV$)

The de-Broglie wavelength of an electron is the same as that of a $50 \ keV$ $X$-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \ MeV$ ).

An electron beam has a kinetic energy equal to $100\; eV$. Find its wavelength (in $\mathring{A}$) associated with the beam,given the mass of an electron $= 9.1 \times 10^{-31}\; kg$,$1\; eV = 1.6 \times 10^{-19}\; J$,and Planck's constant $= 6.6 \times 10^{-34}\; Js$.

The de Broglie wavelength of a proton is twice the de Broglie wavelength of an alpha particle. The ratio of the kinetic energies of the proton and the alpha particle is