The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

What is the de Broglie wavelength (in $\mathring{A}$) of an $\alpha$-particle accelerated through a potential difference of $V$ volts?

The accelerating voltage of an electron gun is $50,000 \ V$. What will be the de Broglie wavelength of the electron in $\mathring{A}$?

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 kinetic energy of an electron gets tripled,then the de-Broglie wavelength associated with it changes by a factor of:

The ratio of the de-Broglie wavelength of an $\alpha$-particle to that of a proton,when both are subjected to the same magnetic field such that the radii of their paths are equal,assuming the magnetic field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton,is: