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 $\frac{K}{4}$?

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

If the de-Broglie wavelengths for a proton and for an $\alpha$-particle are equal, then the ratio of their velocities will be

An electron is released from rest near an infinite non-conducting sheet of uniform charge density $-\sigma$. The rate of change of de Broglie wavelength associated with the electron varies inversely as the $n^{\text{th}}$ power of time. The numerical value of $n$ is . . . . . . .

The de Broglie wavelength of an electron accelerated between two plates having a potential difference of $900 \ V$ is nearly. (in $nm$)

The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?