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

$A$ particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda_x$ and $\lambda_y$ respectively. If $x$ and $y$ were moving in opposite directions,then the de-Broglie wavelength of $P$ is

The de Broglie wavelength and kinetic energy of a particle are $2000 \ \mathring{A}$ and $1 \ \text{eV}$ respectively. If its kinetic energy becomes $1 \ \text{MeV}$,then its de Broglie wavelength becomes $...... \ \mathring{A}$.

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?

Write a note on the electron microscope.

Photons of energy $4.5 \ eV$ are incident on a photosensitive material of work function $3 \ eV$. The de Broglie wavelength associated with the photoelectrons emitted with maximum kinetic energy is nearly (in $Å$)