Question

Calculate the wave number for longest transition in the Balmer series of atomic hydrogen? (its class 11 question)

Answer 1

By the formula for the wave number
109,677(\frac{1}{n _{1} ^{2} } -\frac{1}{n _{2} ^{2} } )
For Balmer series
n1 = 2 and n2 = 3,4,5...
Wavenumber
= 109,699(1/4 - 1/9) = 15232.9  cm^{-1}