Question

The average lifetime of a hydrogen atom excited to the \(n = 2\) state is \(10^{-8}\ \text{s}\). The average number of revolutions the electron makes before it jumps to the ground state is:

• A. \(8.2 \times 10^{6}\)
• B. \(2 \times 10^{6}\)
• C. \(82 \times 10^{6}\)
• D. \(8.2 \times 10^{5}\)

Hint:

Find the Bohr orbital frequency for \(n=2\) (\(f_1/n^3\)) and multiply by the lifetime \(10^{-8}\ \text{s}\).

Answer 1

The Correct Option is A

Step 1: In the Bohr model the orbital frequency scales as \(f_n = \dfrac{f_1}{n^3}\), since \(v_n = v_1/n\) and \(r_n = n^2 r_1\), giving \(f_n = \dfrac{v_n}{2\pi r_n} = \dfrac{f_1}{n^3}\).

Step 2: For the ground state, \(v_1 = 2.19\times10^{6}\ \text{m/s}\) and \(r_1 = 0.529\times10^{-10}\ \text{m}\), so
\[f_1 = \frac{v_1}{2\pi r_1} = \frac{2.19\times10^{6}}{2\pi(0.529\times10^{-10})} \approx 6.59\times10^{15}\ \text{Hz}\]

Step 3: For \(n = 2\):
\[f_2 = \frac{f_1}{2^3} = \frac{6.59\times10^{15}}{8} \approx 8.24\times10^{14}\ \text{Hz}\]

Step 4: The number of revolutions in the lifetime \(\tau = 10^{-8}\ \text{s}\) is
\[N = f_2\,\tau = (8.24\times10^{14})(10^{-8}) \approx 8.2\times10^{6}\]

Step 5: Therefore
\[\boxed{N \approx 8.2\times10^{6}}\]