Energy E of a hydrogen atom with principal quantum number n is given by E=−13.6n2eV. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately

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1.5 eV

0.85 eV

3.4 eV

1.9 eV

answer is D.

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Detailed Solution

E3→2=−3.4−(−1.51)=−1.89eV⇒E3→2≈1.9eV

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