The average life time for then excited state of a hydrogen-like atom is s and that for the state is — Atomic Structure Chemistry Question
Question
The average life time for then $n=3$ excited state of a hydrogen-like atom is $4.8 \times 10^{-8}$ s and that for the $n=2$ state is $1.28 \times 10^{-7}$ s. Ratio of the average number of revolutions made in the $n=3$ state to the average number of revolutions made in the $n=2$ state before any transitions can take place from these states is $1:x$. The value of $x$ is
💡 Solution & Explanation
The number of revolutions $N$ is the total time $t$ divided by the time period of one revolution $T$. The time period $T$ is proportional to $n^3$. Thus, $N \propto \frac{t}{n^3}$. For $n=3$, $N_3 \propto \frac{4.8 \times 10^{-8}}{3^3} = \frac{4.8 \times 10^{-8}}{27}$. For $n=2$, $N_2 \propto \frac{1.28 \times 10^{-7}}{2^3} = \frac{1.28 \times 10^{-7}}{8} = 0.16 \times 10^{-7} = 1.6 \times 10^{-8}$. The ratio is $\frac{N_3}{N_2} = \frac{4.8 / 27}{1.6} = \frac{3}{27} = \frac{1}{9}$. Thus, the ratio is 1:9, so $x = 9$. Therefore, correct answer is 9.