Answer
$7.23 \times 10^{10}$ $rad.s^{-1}$
Work Step by Step
The angular wavenumber is expressed by the expression
$k=\frac{2\pi\sqrt {2m(E-U)}}{h}$ ......................$(1)$,
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 $2$, where $U =U_{1}=600$ $eV$, Eq. $(1)$ tells us that the angular wave number is
$k=\frac{2\pi\sqrt {2\times 9.1\times 10^{-31}\times(800-600)\times1.6\times 10^{-19}}}{6.63\times 10^{-34}}$ $rad.s^{-1}$
or, $k\approx 7.23 \times 10^{10}$ $rad.s^{-1}$
$\therefore$ The angular wave number in region $2$ is $7.23 \times 10^{10}$ $rad.s^{-1}$
