Question

If the average life time of an excited state of H atom is of order $10^{-8}$ sec, estimate how many revolutions (in the order of $10^6$) an $e^-$ makes when it is in the state n = 2 and before it suffers a transition to n =1 state.

• A. 8.2 x 10^6
• B. 9.2 x 10^6
• C. 7.2 x 10^6
• D. 6.2 x 10^6

Answer 1

Answer: (A)

8.2 x 10^6

Answer 2

The frequency of revolution for an electron in the n-th orbit of a hydrogen atom is given by $f_n = \frac{v_n}{2\pi r_n}$. For n=2, the velocity $v_2 = \frac{2.188 \times 10^6}{2}$ m/s and the radius $r_2 = 2^2 \times 0.529 \times 10^{-10}$ m. Thus, $f_2 = \frac{2.188 \times 10^6 / 2}{2\pi \times (4 \times 0.529 \times 10^{-10})} \approx 8.23 \times 10^{14}$ Hz. The number of revolutions is the product of frequency and lifetime: Number of revolutions = $f_2 \times \text{lifetime} = (8.23 \times 10^{14} \text{ s}^{-1}) \times (10^{-8} \text{ s}) \approx 8.23 \times 10^6$.