For a hydrogen atom excited to the n = 2 state, the average...

The time period of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $T_n = \dfrac{2\pi r_n}{v_n}$.

The radius of the $n^{\text{th}}$ orbit is $r_n = 0.529 \times 10^{-10} \times \dfrac{n^2}{Z} \text{ m}$.

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

For a hydrogen atom, $Z = 1$. In the excited state $n = 2$, the radius and velocity are:

$r_2 = 0.529 \times 10^{-10} \times 4 = 2.116 \times 10^{-10} \text{ m}$

$v_2 = \dfrac{2.18 \times 10^6}{2} = 1.09 \times 10^6 \text{ m/s}$

Substituting these values to find the time period $T_2$:

$T_2 = \dfrac{2 \times 3.14 \times 2.116 \times 10^{-10}}{1.09 \times 10^6} \approx 1.22 \times 10^{-15} \text{ s}$

The average lifetime of the electron in the $n = 2$ state is $t = 10^{-8} \text{ s}$.

The average number of revolutions $N$ completed before transitioning to the ground state is the total time divided by the time period:

$N = \dfrac{t}{T_2} = \dfrac{10^{-8}}{1.22 \times 10^{-15}} \approx 8.2 \times 10^6$