Answer
$8.86\times 10^{10}$ $rad.m^{-1}$
Work Step by Step
The angular wave number $(k)$ is related to the de Broglie wavelength $(\lambda)$ by the expression
$k=\frac{2\pi}{\lambda}$
Given $\lambda=7.09\times 10^{-11}$ $m$
$\therefore k=\frac{2\pi}{7.09\times 10^{-11}}$ $rad.m^{-1}$
or, $k=8.86\times 10^{10}$ $rad.m^{-1}$
Therefore, the angular wavenumber of the electron is $8.86\times 10^{10}$ $rad.m^{-1}$.
