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When a hydrogen atom emits a photon of energy 12.1 eV, its orbital angular momentum
changes by

Option: 1

1.05 \times 10^{-34} \mathrm{~J} \mathrm{~s}


Option: 2

2.11 \times 10^{-34} \mathrm{~J} \mathrm{~s}


Option: 3

3.16 \times 10^{-34} \mathrm{~J} \mathrm{~s}


Option: 4

4.22 \times 10^{-34} \mathrm{~J} \mathrm{~s}


Answers (1)

best_answer

Emission of a photon of 12.1 eV requires a transition from n = 3 to n = 1

Change in orbital angular momentum  =\frac{\mathrm{h}}{2 \pi}(3-1)=\frac{\mathrm{h}}{\pi}=2.11 \times 10^{-34} \mathrm{~J} \mathrm{~s}

 

Posted by

Irshad Anwar

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