Discussion
Let \(T_{g}\) and \(T_{e}\) be the kinetic energies of the electron in the ground and the third excited states of a hydrogen atom, respectively. According to the Bohr model, the ratio \(T_{g}/T_{e}\) is
Answer: D
The ratio is 16.
In the Bohr model the kinetic energy of the electron in the nth orbit is \(T_{n} = \frac{13.6}{n^{2}}\) eV, so it falls as \(1/n^{2}\). The ground state is \(n = 1\); the first, second and third excited states are \(n = 2, 3, 4\).
\[ \frac{T_{g}}{T_{e}} = \frac{1/1^{2}}{1/4^{2}} = 16 \]
Taking the third excited state as \(n = 3\) gives 9, the commonest slip. 4 and 3 come from using \(n\) instead of \(n^{2}\).
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