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# Calculate the angular frequency of an electron occupying the second Bohr's orbit of $He^+$ ion.

$\begin{array}{1 1}(a)\;2.067\times 10^{16}sec^{-1}&(b)\;1.32\times 10^{16}sec^{-1}\\(c)\;2.067\times 10^2sec^{-1}&(d)\;3.2\times 10^{20}sec^{-1}\end{array}$

Velocity of electron in $He^+$ ion u=$\large\frac{2\pi ze^2}{nh}$
Radius of $He^+$ ion in an orbit $(r_n)=\large\frac{n^2h^2}{4\pi^2me^2z}$
$\therefore$ Angular frequency or angular velocity $\omega$
$\omega=\large\frac{u}{r_n}$
$\Rightarrow \large\frac{2\pi z e^2\times 4\pi^2me^2z}{nh\times n^2h^2}$
$\Rightarrow \large\frac{8\pi^3z^2me^4}{n^3h^3}$
$n=2,m=9.108\times 10^{-28}g,z=2,h=6.625\times 10^{-27}$
$\therefore \omega=\large\frac{8\times (22/7)^3\times 2^2\times 9.108\times 10^{-28}\times (4.803\times 10^{-10})^4}{2^3\times (6.625\times 10^{-27})^3}$
$\Rightarrow 2.067\times 10^{16}sec^{-1}$
Hence (a) is the correct answer.