λ1=RZ2(n121−n221)
λ11=R(1)2(221−421)=163R
λ1λ2=2720
λ2=2720×6561A∘=4860A∘
=486nm
JEE Main 2020 — Physics Modern Physics
The first member of the Balmer series of hydrogen atom has a wavelength of 6561A∘. The wavelength of the second member of the Balmer series (in nm) is_____________
Held on 8 Jan 2020 · Verified 6 Jul 2026.
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