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Chapter 19: Problem 62

An electron in hypothetical hydrogen atom is in its \(3^{\text {rd }}\) excited state and makes transition from \(3^{\text {rd }}\) to \(2^{\text {nd }}\) excited, then to \(1^{\text {st }}\) excited state and then to ground state. If the amount of time spent by the electron in any state of quantum number \(n\), is proportional to \(\left(\frac{1}{n-1}\right)\), then the ratio of number of revolutions completed by the electron in \(1^{\text {st }}\) excited state to that in the \(2^{\text {nd }}\) excited state will be (A) 2 (B) \(\frac{27}{8}\) (C) \(\frac{27}{4}\) (D) \(\frac{27}{6}\)

Short Answer

Expert verified
The ratio of the number of revolutions completed by the electron in the \(1^{\text {st }}\) excited state to that in the \(2^{\text {nd }}\) excited state is 2 (A).

Step by step solution

01

Calculate the time spent in each excited state

According to the problem, the time spent by the electron in a state with quantum number \(n\) is proportional to \(\left(\frac{1}{n-1}\right)\). For the \(1^{\text {st }}\) excited state, \(n=2\): \(time_1 \propto \left(\frac{1}{2-1}\right) = 1\) For the \(2^{\text {nd }}\) excited state, \(n=3\): \(time_2 \propto \left(\frac{1}{3-1}\right) = \frac{1}{2}\) Now, we'll use these proportions to find the ratio of the times spent in the excited states: \(\frac{time_1}{time_2}\)
02

Find the ratio of the times spent in the excited states

To find the ratio of the times spent in the excited states, we divide \(time_1\) by \(time_2\): \(\frac{time_1}{time_2} = \frac{1}{\frac{1}{2}} = 2\) This means that the electron spends twice as much time in the \(1^{\text {st }}\) excited state compared to the \(2^{\text {nd }}\) excited state.
03

Find the ratio of the number of revolutions in the excited states

The number of revolutions completed by the electron in each excited state is directly proportional to the time spent in that state. Since we have already determined that the electron spends twice as much time in the \(1^{\text {st }}\) excited state compared to the \(2^{\text {nd }}\) excited state, the ratio of the number of revolutions completed in each state is also 2. Hence, the ratio of the number of revolutions completed by the electron in the \(1^{\text {st }}\) excited state to that in the \(2^{\text {nd }}\) excited state is 2, which is the correct answer (A).

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