Proton $(P)$ and electron $(e)$ will have the same de Broglie wavelength when the ratio of their momentum is (assume,$m_{p} = 1849 \, m_{e}$)

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

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

If ${\lambda _p}$ and ${\lambda _\alpha }$ are the de-Broglie wavelengths associated with protons and $\alpha$-particles of equal kinetic energies,then:

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 associated with a proton under the influence of an electric potential of $100 \ V$ is