12.7 (a) Using the Bohr’s 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.

Hint: The centripetal force to the electron is provided by the electrostatic force between the nucleus and the electron.
(a)
Step:
Find the velocity of the electron in the first orbit.
Let ${\mathrm{v}}_{1}$ be the orbital speed of the electron in a hydrogen atom in the ground-state level, ${\mathrm{n}}_{1}$ =1.
For an electron, is given by the relation,
$$v_{1}=\frac{e^{2}}{n_{1} 4 \pi \epsilon_{0} \frac{h}{2 \pi}}=\frac{e^{2}}{2 \epsilon_{0} h}$$
where,
${\in }_{0}$ = Permittivity of free space =
h = Planck's constant =
$${v}_{1}=\frac{1.6 \times 10^{-19^{2}}}{2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}=0.0218 \times 10^{8}=2.18 \times 10^{6} \mathrm{~m} / \mathrm{s}$$
For level ${\mathrm{n}}_{2}$ =  2, we can write the relation for the corresponding orbital speed as:
$$v_{2}=\frac{e^{2}}{n_{2}\times2 \epsilon_{0} h}=\frac{n_1}{2}=1.09 \times 10^{6} \mathrm{~m} / \mathrm{s}$$
And, for ${\mathrm{n}}_{3}$ = 3, we can write the relation for the corresponding orbital speed as
$$v_{3}=\frac{e^{2}}{n_{3}\times2 \epsilon_{0} h}=\frac{n_1}{3}=7.27 \times 10^{5} \mathrm{~m} / \mathrm{s}$$
(b)
Hint: $$T=\frac{2\pi~r}{v}$$
Step:
Find the time period of the electron in the first orbit.
Let ${\mathrm{T}}_{1}$ be the orbital period of the electron when it is in level ${\mathrm{n}}_{1}=1.$
Orbital period is related to orbital speed as:
${\mathrm{T}}_{1}=\frac{2{\mathrm{\pi r}}_{1}}{{\mathrm{v}}_{1}}$
where, ${\mathrm{r}}_{1}$  = Radius of the orbit=$$\frac{n_{1}^{2} h^{2} \epsilon_{0}}{\pi m e^{2}}$$
h = Planck's constant =  and
e = Charge on an electron
${\in }_{0}$ = Premitivity of free space $=8.85×{10}^{-12}{\mathrm{N}}^{-1}{\mathrm{C}}^{2}{\mathrm{m}}^{-2}$
m = mass of an electron $=9.1×{10}^{-31}\mathrm{kg}$
$$T_{1}=\frac{2 \pi r_{1}}{v_{1}}=\frac{2 \pi \times 1^{2} \times 6.62 \times 10^{-34^{2}} \times 8.85 \times 10^{-12}}{2.18 \times 10^{6} \times \pi \times 9.1 \times 10^{-31} 1.6 \times 10^{-19^{2}}}=15.27 \times 10^{-17}=1.527 \times 10^{-16} \mathrm{sec}$$
$$T_{2}=\frac{2 \pi r_{2}}{v_{2}}={n_2}^{3} \times (T_1)=1.22\times10^{-15}~sec$$
$$T_{3}=\frac{2 \pi r_{3}}{v_{3}}={n_3}^{3} \times (T_1)=4.12\times10^{-15}~sec$$