Answer
$2.486\times 10^{-8}$ $nm$
Work Step by Step
As the energy ($E$) of the electron is very large, we can assume the extreme relativistic relation between $E$ and momemtum $p$:
$p=\frac{E}{c}$ ................$(1)$
In this extreme situation, the kinetic energy ($K$) of the electron is much greater than its rest energy.
Now, The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$ ......................$(2)$
where $h$ is the planck's constant
Substituting eq. $1$ in eq. $2$, we get
$λ=\frac{hc}{E}$ ......................$(3)$
Here, $E=50$ $GeV$,
Thus, the corresponding de Broglie wavelength is given by
$λ=\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{50\times 10^9\times 1.6\times 10^{-19}}$ $m$
or, $λ=2.486\times 10^{-17}$ $m$
or, $λ=2.486\times 10^{-8}$ $nm$
