The ratio of the de-Broglie wavelength of an $\alpha$-particle to that of a proton,when both are subjected to the same magnetic field such that the radii of their paths are equal,assuming the magnetic field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton,is:

The de Broglie wavelength of a proton accelerated by a potential difference of $100 \ V$ is $\lambda_0$. If an alpha particle is accelerated by the same potential difference,its de Broglie wavelength will be:

The wavelength of a very fast-moving electron $(v \approx c)$ is:

$A$ particle is moving along the $x-$axis back and forth in a box of length $L$. Assuming the de-Broglie hypothesis is applicable for the particle,and it is moving with a constant speed in the box making perfectly elastic collisions with the walls. The possible value of the momentum of the particle is (Note: $h$ is Planck's constant,$n$ is the principal quantum number).

What is the experimental value of the de-Broglie wavelength for an electron accelerated through a potential difference of $V$ volts?

An electron and a photon each have a wavelength of $1.00 \; nm$. Find
$(a)$ their momenta,
$(b)$ the energy of the photon,and
$(c)$ the kinetic energy of the electron.