Q.Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.

[!TLDR] Since $v_n\propto1/n$ and $r_n\propto n^2$, angular speed $\omega_n=v_n/r_n\propto1/n^3$ -- maximum at the smallest allowed n, i.e. the ground state (n=1). For hydrogen this gives $\omega_1=v_1/r_1\approx4.13\times10^{16}$ rad/s. [!ANSWER] Maximum angular speed occurs at n = 1 (ground state), $\omega_{max}\approx4.13\times10^{16}$ rad/s.

Angular speed is $\omega_n=v_n/r_n$. Using the Bohr-model results $v_n=\frac{Ze^2}{2\epsilon_0hn}\propto\frac{1}{n}$ and $r_n=\frac{n^2h^2\epsilon_0}{\pi m_eZe^2}\propto n^2$, the angular speed is $\omega_n=\frac{v_n}{r_n}\propto\frac{1/n}{n^2}=\frac{1}{n^3}$. Since $\omega_n$ falls off as $1/n^3$, it is LARGEST at the smallest allowed value of n, which is n=1 (the ground state) -- there is no smaller, faster orbit available; the electron simply cannot occupy n=0. For hydrogen (Z=1), the ground-state orbit has $v_1\approx2.188\times10^6$ m/s and $r_1=a_0=5.3\times10^{-11}$ m, giving $\omega_1=\frac{v_1}{r_1}=\frac{2.188\times10^6}{5.3\times10^{-11}}\approx4.128\times10^{16}$ rad/s -- this is the maximum angular speed any electron can have in a stable orbit of a hydrogen atom. [!ANSWER] Maximum angular speed occurs at n = 1 (ground state), $\omega_{max}\approx4.13\times10^{16}$ rad/s.

Combine $v_n\propto1/n$ and $r_n\propto n^2$ into $\omega_n=v_n/r_n\propto1/n^3$, identify that this is maximised at the smallest allowed n (=1), then substitute the known ground-state values.

Assuming angular speed increases with n (confusing it with the growing ORBIT SIZE or with linear speed's own trend), when in fact both linear speed and angular speed DECREASE as n increases -- only the radius grows.