Problem

Problem

The energy of the n=2n = 2 state in a given element is E2=−2870E_2 = -2870 eV. Given that the wavelengths of the KαK_\alpha and KβK_\beta -lines are 0.71 nm and 0.63 nm respectively, determine the energies E1E_1 and E3E_3 .

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