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Answer :
For a hydrogen atom, the excited state (n2) if a wavelength of 97.3 nm is emitted when n1 = 1 is n2 = 3.
In a hydrogen atom, the energy levels of the electron are given by En = -13.6/n² eV, where n is the principal quantum number. This formula gives the energy levels of the hydrogen atom when the electron is in its ground state (n=1).When an electron in an atom jumps from a higher energy level to a lower energy level, a photon is emitted, and the energy of the photon is equal to the difference between the two energy levels (ΔE).
The wavelength of the photon emitted using the formula:hc/λ = ΔEwhere h is Planck's constant (6.626 × 10⁻³⁴ J.s), c is the speed of light (2.998 × 10⁸ m/s), and λ is the wavelength of the emitted photon.So, if a wavelength of 97.3 nm is emitted when n1 = 1, we can calculate the energy difference (ΔE) between the energy level of the electron in the excited state (n2).
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