λ11=R(321−421)=144R×7
λ21=R(221−321)=36R×5
⇒λ2λ1=36R×5×R×7144=720
JEE Main 2019 — Physics Modern Physics
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths λ2λ1 of the photons emitted in this process is:
Held on 12 Apr 2019 · Verified 6 Jul 2026.
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720
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527
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