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If a hydrogen atom emits a photon of energy 12.75 eV and returns to its ground state, change in its orbital angular momentum is

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detailed solution

Correct option is A

(1)ΔE=13.6 1−1n2∴ 12.75 = 13.6 1−1n2 ⇒ n=4∴ ΔL=4.hπ−1.h2π=3h2π

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