Question

what is the frequency of electron in second Bohr's orbit ?

Answer 1

The n-th Bohr's orbit radius is

$$R _ {n} = \frac {4 \pi \varepsilon_ {0} \hbar^ {2}}{Z m _ {e} e ^ {2}} n ^ {2},$$

where the $\varepsilon_0 = 8.85 \times 10^{-12} F / m$ is the vacuum permittivity, $\hbar = 1.05 \times 10^{-34} J \cdot s$ is the Plank constant, $n$ is the number of orbit ( $n = 2$ in the our case), $Z$ is the charge of the atom's core ( $Z = 1$ for the atom of Hydrogen), $m_e = 9.1 \times 10^{-31} kg$ is the mass of electron and $e = 1.6 \times 10^{-19} C$ is the modulus of the electron charge. For the stationary orbit the following condition is met:

$$m _ {e} v R _ {n} = n \hbar ,$$

where $v$ is the electron's speed. One can calculate the frequency as

$$f = \frac {1}{T} = \frac {v}{2 \pi R _ {n}}.$$

Thus substituting the expressions for $R_{n}$ from the first formula and for $v$ from the second one, one obtains the expression for the frequency

$$f = \frac {Z ^ {2} m _ {e} e ^ {4}}{3 2 \pi^ {3} \varepsilon_ {0} ^ {2} \hbar^ {3} n ^ {3}}.$$

For the Hydrogen one gets $f = 8.3 \, \text{THz}$ .