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What will be the longest wavelength line in Balmer series of spectrum of hydrogen atom?

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For single electron species,

\(\frac{1}{λ} = RZ^2 (\frac{1}{n^2_1} - \frac{1}{n^2_2})\)

where, R = Rydberg constant = 1.097×107×10−9nm−1

= 1.097×10−2nm−1

For Hydrogen atom, Z = Atomic number = 1

For Balmer series, n1​ = 2 and for longest wavelength in Balmer series, minimum energy transition is to be considered because wavelength is inversely proportional to Energy.

So, n2​ = 3

\(\frac{1}{λ}\)​= 1.097×10−2nm−1×12\((\frac{1}{2^2}-\frac{1}{3^2})\)

⇒ λ = 656nm

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