Answer
$1.24$ $fm$
Work Step by Step
As the energy (E) of the electron is very large, we can assume the extreme relativistic relation between E and momentum 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 ($λ$) is express by the formula,
$λ=\frac{h}{p}$ ......................($2$)
where h is the planck's constant
Substituting eq. $1$ in eq. $2$, we get
$λ=\frac{hc}{E}$ ......................($3$)
Here, $E=1.0$ $GeV$,
Thus, the corresponding de Broglie wavelength of the electron is given by
$λ=\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{1.0\times 10^{9}\times 1.6\times 10^{-19}}$ $m$
or, $λ\approx1.24\times 10^{-15}$ $m$
or, $λ=1.24$ $fm$
