From Rydberg’s equation,
λ1=R(n21−n221)
660×10−91=R(221−321)
λ1=R(221−421)
600×10−9λ=9×45×1216×4
λ=488.9nm
JEE Main 2019 — Physics Modern Physics
Taking the wavelength of first Balmer line in hydrogen spectrum (n=3 to n=2) as 660nm , the wavelength of the 2nd Balmer line (n=4 to n=2) will be :
Held on 9 Apr 2019 · Verified 6 Jul 2026.
889.2nm
488.9nm
388.9nm
642.7nm
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