If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is

$A$ photon and an electron have equal energy $E$. The ratio of $\lambda(\text{electron})$ to $\lambda(\text{photon})$ is proportional to

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

The de Broglie wavelengths associated with a proton and an electron are in the ratio $2: 1$. Their stopping potentials are approximately in the ratio of

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

The de Broglie wavelength associated with a neutron at temperature $T$ is given by: $(E = kT)$