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: The revolution count is the orbital frequency in the $n=2$ state multiplied by the time spent there. Write the frequency in terms of the ground-state period.

Step 2: The Bohr period is $T_n = \dfrac{2\pi r_n}{v_n} = T_1 n^3$, because radius grows as $n^2$ and speed falls as $1/n$. The ground-state period is $T_1 = \dfrac{2\pi r_1}{v_1} = \dfrac{2\pi(0.529\times10^{-10})}{2.19\times10^{6}} \approx 1.52\times10^{-16}\ \text{s}$.

Step 3: For $n=2$ the period is $T_2 = 2^3 T_1 = 8(1.52\times10^{-16}) \approx 1.21\times10^{-15}\ \text{s}$.

Step 4: The number of orbits completed in the lifetime $\tau = 10^{-8}\ \text{s}$ is
\[N = \frac{\tau}{T_2} = \frac{10^{-8}}{1.21\times10^{-15}} \approx 8.2\times10^{6}\]

Step 5: So the electron circles about eight million times before decaying:
\[\boxed{N \approx 8.2\times10^{6}}\]