An electron accelerated through a potential of $10000 \ V$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential,so that the wavelength is doubled (in $V$)?

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

$A$ proton and an $\alpha-$particle are accelerated through a potential difference of $100 \ V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha-$particle is:

$A$ particle has a de Broglie wavelength $\lambda$. If its velocity and energy are both doubled,what will be the new de Broglie wavelength?

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

An electron with speed $v$ and a photon with speed $c$ have the same $de-Broglie$ wavelength. If the kinetic energy and momentum of the electron are $E_{e}$ and $p_{e}$ and that of the photon are $E_{ph}$ and $p_{ph}$ respectively,which of the following is correct?