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What is the wavelength of the photon emitted when an electron in a hydrogen atom relaxes from the second excited state to the ground state using quantum chemical calculations of excited states?

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To calculate the wavelength of the photon emitted when an electron in a hydrogen atom relaxes from the second excited state to the ground state, we can use the Rydberg formula for hydrogen:1/ = R_H *  1/n1 - 1/n2 where  is the wavelength of the emitted photon, R_H is the Rydberg constant for hydrogen  approximately 1.097 x 10^7 m , n1 is the principal quantum number of the lower energy level  ground state , and n2 is the principal quantum number of the higher energy level  second excited state .For hydrogen, the ground state corresponds to n1 = 1, and the second excited state corresponds to n2 = 3.1/ = R_H *  1/1 - 1/3 1/ = 1.097 x 10^7 m *  1 - 1/9 1/ = 1.097 x 10^7 m *  8/9 Now, we can calculate the wavelength: = 1 /  1.097 x 10^7 m * 8/9   1.215 x 10 mSo, the wavelength of the photon emitted when an electron in a hydrogen atom relaxes from the second excited state to the ground state is approximately 1.215 x 10 m or 121.5 nm. This photon is in the ultraviolet range of the electromagnetic spectrum.
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