Question

Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen

Answer 1

Correct Answer - A::B::C::D
`bar v = (1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)`
For the shortest wavelength of Balmer series
`n_(1) = 2 and n_(2) = prop (H atom)`
`bar v = 109678((1)/(2^(2)) - (1)/(oo^(2)))1^(2)`
`= 27419.25 cm^(-1)`