$(a)$ Using the Bohr model,calculate the speed of the electron in a hydrogen atom in the $n = 1, 2,$ and $3$ levels.
$(b)$ Calculate the orbital period in each of these levels.

(N/A) Let $v_{n}$ be the orbital speed of the electron in a hydrogen atom in the $n$-th level. The speed is given by $v_{n} = \frac{v_{1}}{n}$,where $v_{1} = \frac{e^{2}}{2 \epsilon_{0} h} \approx 2.18 \times 10^{6} \, m/s$.
For $n=1$: $v_{1} = 2.18 \times 10^{6} \, m/s$.
For $n=2$: $v_{2} = \frac{v_{1}}{2} = 1.09 \times 10^{6} \, m/s$.
For $n=3$: $v_{3} = \frac{v_{1}}{3} = 7.27 \times 10^{5} \, m/s$.
$(b)$ The orbital period $T_{n}$ is given by $T_{n} = \frac{2 \pi r_{n}}{v_{n}}$. Since $r_{n} = n^{2} r_{1}$ and $v_{n} = \frac{v_{1}}{n}$,we have $T_{n} = n^{3} T_{1}$,where $T_{1} = \frac{2 \pi r_{1}}{v_{1}} \approx 1.52 \times 10^{-16} \, s$.
For $n=1$: $T_{1} = 1.52 \times 10^{-16} \, s$.
For $n=2$: $T_{2} = 2^{3} T_{1} = 8 \times 1.52 \times 10^{-16} = 1.22 \times 10^{-15} \, s$.
For $n=3$: $T_{3} = 3^{3} T_{1} = 27 \times 1.52 \times 10^{-16} = 4.12 \times 10^{-15} \, s$.