Answer
$7.09\times 10^{-11}$ $m$
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.
Here, $p=9.35\times 10^{-24}$ $kg.m.s^{-1}$
$\therefore\lambda=\frac{6.63\times 10^{-34}}{9.35\times 10^{-24}}$ $m$
or, $\lambda\approx7.09\times 10^{-11}$ $m$
Therefore, the de Broglie wavelength of the electron is $7.09\times 10^{-11}$ $m$.
