The first member of the Balmer series of hydrogen atom has a wavelength of 6561   A ∘ . The…

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A∘. The wavelength of the second member of the Balmer series (in nm) is_____________

Solution

1λ=RZ21n12-1n22
1λ1=R12122-142=3R16
λ2λ1=2027
λ2=2027×6561 A∘=4860 A∘
=486 nm

Asked in: JEE Main 2020 (08 Jan Shift 2)

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