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(C) The total energy of an electron in a hydrogen atom is given by the formula:$E_{n} = \frac{-13.6}{n^{2}} 	ext{ eV}$From this,we can see that the e

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$4: 1$

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(C) The total energy of an electron in a hydrogen atom is given by the formula:$E_{n} = \frac{-13.6}{n^{2}} 	ext{ eV}$From this,we can see that the energy is inversely proportional to the square of the principal quantum number:$E_{n} \propto \frac{1}{n^{2}}$The first excited state corresponds to $n = 2$,and the third excited state corresponds to $n = 4$.Therefore,the ratio of the energy in the first excited state $(E_{2})$ to the energy in the third excited state $(E_{4})$ is:$\frac{E_{2}}{E_{4}} = \frac{n_{4}^{2}}{n_{2}^{2}} = \left(\frac{4}{2}ight)^{2}$$\frac{E_{2}}{E_{4}} = (2)^{2} = \frac{4}{1}$Thus,the ratio is $4: 1$.

What is the ratio of the total energy of an electron in a hydrogen atom in the first excited state to that in the third excited state?

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