Solution:
The orbital period is proportional to \(n^3\). Since \(T_1 = 8 T_2\), we must have \(n_1^3 = 8 n_2^3\), which gives \(n_1 = 2n_2\). Thus, \(n_1 = 4\) and \(n_2 = 2\) is correct.
An electron in a hydrogen atom makes a transition from \(n = n_1\) to \(n = n_2\). The time period of revolution of the electron in the initial state is eight times that in final state. The possible value of \(n_1\) and \(n_2\) are
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