The de-Broglie wavelength $\lambda$ associated with an electron having kinetic energy $E$ is given by the expression:

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 $0.5 \ MeV$ ).

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

Find the de Broglie wavelength of an electron in a metal at $27^{\circ}C$.

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be: