Answer
$7.72\times 10^{10}$ $m^{-1}$
Work Step by Step
The angular wavenumber is expressed by the expression
$k=\frac{2\pi\sqrt {2m(E-U)}}{h}$,
in which $E$ and $U$ are the total mechanical and potential energy of the moving particle respectively. $m$ is the mass of the particle. $h$ is planck's constant.
In region $2$,
the value of electric potential is $V_{2}=-500$ $V$
$U=500$ $eV$
$E=728$ $eV$
$E-U=228$ $eV$
Thus, in region $2$,
$k=\frac{2\pi\sqrt {2\times 9.1\times 10^{-31}\times 228\times 1.6\times 10^{-19}}}{6.63\times 10^{-34}}$ $m^{-1}$
$k\approx 7.72\times 10^{10}$ $m^{-1}$
$\therefore$ The value of angular wavenumber in region $2$ is $7.72\times 10^{10}$ $m^{-1}$
