Answer
$1.24\times 10^{9}$ $eV$
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
$\therefore$ $p=\frac{h}{\lambda}$ ......................$(1)$
The de Broglie wavelength of the electron is given: $\lambda=1.00$ $fm$. For shorter de Broglie wavelength, we have to consider relativistic approach to find the kinetic energy ($K$) of an electron
Total energy ($E$) of an electron is expressed as
$E=\sqrt {p^2c^2+m^2c^4}$
and rest mass energy of and electron is given by
$E_0=mc^2$
Now, the kinetic energy $K$ of the electron is given by
$K=E-E_0$
or, $K=\sqrt {p^2c^2+m^2c^4}-mc^2$
Substituting eq. $1$ in the above relation, we get
$K=\sqrt {\frac{h^2c^2}{\lambda^2}+m^2c^4}-mc^2$
Substituting $\lambda=1.00$ $fm$, we get
$K=\sqrt {\frac{(6.63\times 10^{-34}\times 3\times 10^{8})^2}{(1\times 10^{-15})^2}+(9.1\times 10^{-31})^2\times(3\times 10^{8})^4}-9.1\times 10^{-31}\times(3\times 10^{8})^2 J$
$K\approx1.988\times 10^{-10}$ $J$
or, $K=\frac{1.988\times 10^{-10}}{1.6\times 10^{-19}}$ $eV$
or, $K\approx 1.24\times 10^{9}$ $eV$
$\therefore$ The kinetic energy of an electron with de Broglie wavelength $1.00$ $fm$ is $1.24\times 10^{9}$ $eV$
