Which of the transitions in hydrogen atom emits a photon of lowest frequency (n = quantum number)?

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n = 2 to n = 1

n = 4 to n = 3

n = 3 to n = 1

n = 4 to n = 2

answer is B.

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Detailed Solution

From diagramE1=−13.6−−3.4=−10.2 eVE2=−13.6−−1.51=−12.09 eVE3=−1.51−−0.85=−0.66 eVE4=−3.4−−1.51=−1.89 eVE3 is least i.e., frequency is lowest

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