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

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperatures $27^oC$ and $127^oC$ respectively is

The ratio of the momentum of an electron to that of an alpha particle, both accelerated from rest through the same potential difference $V$, is:

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$,their de Broglie wavelength in the frame of reference attached to their centre of mass is

$A$ proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ are the kinetic energies of the proton and electron respectively,then choose the correct relation:

The wavelength associated with an electron accelerated through a potential difference of $100 \ V$ is nearly .............. $ \mathring A $