Determine the maximum angular speed of the electron in a stationary...

The angular speed of an electron in the $n$-th orbit of a hydrogen atom is given by $\omega_n = \dfrac{v_n}{r_n}$.

The velocity of the electron in the $n$-th orbit is $v_n = \dfrac{v_1}{n}$, where $v_1 \approx 2.18 \times 10^6 \text{ m/s}$.

The radius of the $n$-th orbit is $r_n = r_1 n^2$, where $r_1 \approx 0.529 \times 10^{-10} \text{ m}$.

Substituting these into the expression for angular speed gives $\omega_n = \dfrac{v_1}{r_1 n^3}$.

For the angular speed to be maximum, the principal quantum number $n$ must be minimum, which is $n = 1$ for the ground state.

Substituting $n = 1$, the maximum angular speed is \omega_\max = \dfrac{v_1}{r_1}.

\omega_\max = \dfrac{2.18 \times 10^6}{0.529 \times 10^{-10}} \approx 4.12 \times 10^{16} \text{ rad/s}.

Rounding to the given options, we get $4.1 \times 10^{16} \text{ rad/s}$.