The $log-log$ graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be

An $\alpha$-particle,a proton,and an electron have the same kinetic energy. Which one of the following is correct in the case of their De-Broglie wavelength?

An electron (mass $m$) with an initial velocity $v = v_0 \hat{i}$ is in an electric field $E = E_0 \hat{j}$. If $\lambda_0 = h/mv_0$,its de Broglie wavelength at time $t$ is given by

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ 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

Photons of energy $4.5 \ eV$ are incident on a photosensitive material of work function $3 \ eV$. The de Broglie wavelength associated with the photoelectrons emitted with maximum kinetic energy is nearly (in $Å$)