Final answer:
The energy of an electron in the second excited state of a hydrogen atom is calculated using the formula En = -13.6 eV/n². For n=3, it yields -1.51 eV, which is not listed among the provided options. If a mistake is made and the 'second excited state' is misconstrued as n=2, the energy would then be -3.4 eV.
Step-by-step explanation:
The energy of the electron in the second excited state of a hydrogen atom can be calculated using the formula for the energy levels of a hydrogen atom: En = -13.6 eV/n². The second excited state corresponds to n = 3, since the first excited state is n = 2. Plugging this into the formula gives us the energy for this state: E3 = -13.6 eV/3² = -13.6 eV/9 = -1.51 eV.
However, this is not one of the options provided. The closest matching and correct answer for this is -1.51 eV, which is not listed. None of the provided options are correct for the second excited state (n=3), but if you consider the 'second excited state' to mean n=2 (which would traditionally be called the 'first excited state'), the energy is E2 = -13.6 eV/2² = -13.6 eV/4 = -3.4 eV, which matches option (a).