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Energy E of a hydrogen atom with principal quantum number n is given by E = −13.6/n² eV. The energy of photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately

1. 1.9 eV
2. 1.5 eV
3. 0.85 eV
4. 3.4 eV

Correct answer: 3.4 eV

Solution

The energy difference between the two states is calculated as ΔE = E3 - E2 = (-13.6/3²) - (-13.6/2²) = -1.51 eV - (-3.4 eV) = 1.89 eV, which is approximately 1.9 eV.

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