The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$. What will be the wavelength of the particle if its kinetic energy becomes $\frac{K}{4}$?

An electron of mass $m_{e}$ and a proton of mass $m_{p} = 1836 m_{e}$ are moving with the same speed. The ratio of their de Broglie wavelength $\frac{\lambda_{\text{electron}}}{\lambda_{\text{proton}}}$ will be ....... .

If an electron and a proton have the same de-Broglie wavelength,then the kinetic energy of the electron 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).

Find the de Broglie wavelength of an electron in a metal at $27^{\circ}C$.

What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$?