Answer
$1.35\times 10^{11}$ $m^{-1}$
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 $1$,
$U=0$
$E-U=$ Kinetic energy of the electron
or, $E-U=\frac{1}{2}\times 9.1\times 10^{-31}\times (1.60\times 10^{7})^2$ $J$
or, $E-U=\frac{\frac{1}{2}\times 9.1\times 10^{-31}\times (1.60\times 10^{7})^2}{1.6\times 10^{19}}$ $eV$
or, $E-U=728$ $eV$
Thus, in region $1$,
$k=\frac{2\pi\sqrt {2\times 9.1\times 10^{-31}\times 728\times 1.6\times 10^{-19}}}{6.63\times 10^{-34}}$ $m^{-1}$
$k\approx 1.35\times 10^{11}$ $m^{-1}$
$\therefore$ The value of angular wavenumber in region $1$ is $1.35\times 10^{11}$ $m^{-1}$
