Answer
$1.51$ $eV$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$
where $h$ is the planck's constant and $p$ is the momentum of the moving particle
$\therefore$ $p=\frac{h}{\lambda}$ ......................$(1)$
Now, the relation between $p$ and kinetic energy $K$ of an electron is expressed as
$K=\frac{p^2}{2m}$ , where $m$ is the mass of the electron
Substituting eq. $1$ in above relation, we get the kinetic energy of an electron
$K=\frac{h^2}{2m\lambda^2}$
The de Broglie wavelength of the electron is given: $\lambda=1.00$ $nm$
$\therefore$ The kinetic energy of the electron is given by
$K=\frac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times (1\times 10^{-9})^2}$ $J$
or, $K=\frac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times (1\times 10^{-9})^2\times 1.6\times 10^{-19}}$ $eV$
or, $K\approx1.51$ $eV$
