The de Broglie wavelength $\lambda$ associated with a proton increases by $25 \%$,if its momentum is decreased by $p_0$. The initial momentum was

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

What is the de Broglie wavelength of an electron accelerated through $80 \ V$ in $\mathring A$?

$A$ light of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. The photoelectrons emitted from the surface have de-Broglie wavelength $\lambda_1$. Then the ratio $\lambda : \lambda_1^2$ is ($h =$ Planck's constant,$c =$ velocity of light,$m =$ mass of electron).

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is:

The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100 \ V$ is