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

$A$ particle having electric charge $3 \times 10^{-19} \text{ C}$ and mass $6 \times 10^{-27} \text{ kg}$ is accelerated by applying an electric potential of $1.21 \text{ V}$. The wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12} \text{ m}$. The value of $\alpha$ is . . . . . . . (Take Planck's constant $h = 6.6 \times 10^{-34} \text{ J} \cdot \text{s}$)

The de-Broglie wavelength of a body of mass $m$ and kinetic energy $E$ is given by:

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

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .