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According to Bohr’s theory, the energy required for an electron in the Li²⁺ ion to be emitted from n = 2 state is (given that the ground state ionization energy of hydrogen atom is 13.6 eV):

1. 61.2 eV
2. 13.6 eV
3. 30.6 eV
4. 10.2 eV

Correct answer: 30.6 eV

Solution

The energy of an electron in a hydrogen-like ion is given by En = -13.6 Z²/n² eV. For Li²⁺ (Z = 3) in the n = 2 state, En = -13.6 × 3² / 2² = -30.6 eV. To emit the electron, 30.6 eV is required to bring it to zero energy.

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