Asked by hellen
An electron in a hydrogen atom makes a transition from an energy state of principal quantum numbers ni to n=2.If the photon emitted has a wavelentgh of 434nm what is the value of ni?
Answers
Answered by
bobpursley
energyNi=energyN2+energy Photon
1.602e-19/Ni^2=1.602e-19/2^2 + 6.626e–34* 3e8/434e-9
solve for Ni.
1.602e-19/Ni^2=1.602e-19/2^2 + 6.626e–34* 3e8/434e-9
solve for Ni.
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