A spectral line of Lyman series of H -atom has a frequency of 2.466 \times 10^{15} { s}^{-1}. What is the transition responsible for this spectral line? ( {\R}=1.096 \times 10^7 {\m}^{-1})

Correct Answer is : n2=2 to n1=1

AP EAPCET Medical 19 May 2025 Shift 1 Paper

Question : 121 of 160

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A spectral line of Lyman series of H -atom has a frequency of 2.466×1015 s−1. What is the transition responsible for this spectral line?
(R=1.096×107m−1)

‌n2=2‌ to ‌n1=1
‌n2=3‌ to ‌n1=1
‌n2=4‌ to ‌n1=2
‌n2=5‌ to ‌n1=1

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