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If a hydrogen atom emit a photon of energy `12.1 eV` , its orbital angular momentum changes by `Delta L. then`Delta L` equalsA. `1.05xx10^(-34)J" "sec`B. `2.11xx10^(-34)J" "sec`C. `3.16xx10^(-34)J" "sec`D. `4.22xx10^(-34)J" "sec` |
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Answer» Correct Answer - B Emission of photo of 12.1 eV corresponds to the `therefore` Change in angular momantum `=(n_(2)-n_(1))(h)/(2pi)` `=(3-1)(h)/(2pi)=(h)/(pi)` `=(6.626xx10^(-34))/(3.14)=2.11xx10^(-34)J" "sec` |
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