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The energy required to ionize a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

1. 11.4 nm
2. 24.2 nm
3. 8.6 nm
4. 35.8 nm

Correct answer: 11.4 nm

Solution

Ionization energy = 9 Ry gives Z^2 = 9. Energy of n=3 -> n=1 transition = Z^2 * 13.6 * (1 - 1/9) = 9*13.6*(8/9) = 108.8 eV. Wavelength = 1240/108.8 = 11.4 nm.

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