Question

The frequency of one of the lines in Paschen series of hydrogen atom is `2.340 xx 10^(14)Hz`. What is the quantum number `n_(2)` which produces this transition ?

Answer 1

`bar(v) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`bar(v) = (1)/(lamda), c = v lamda or (1)/(lamda) = (v)/(c)`
`:. v = c R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
For Paschen series, `n_(1) = 3`
`:. 2.340 xx 10^(14) = (3 xx 10^(8) ms^(-1)) xx (1.097 xx 10^(7) m^(-1)) ((1)/(3^(2)) - (1)/(n_(2)^(2)))`
or `((1)/(3^(2)) - (1)/(n_(2)^(2))) = (2.340 xx 10^(14))/(3.291 xx 10^(15)) = 0.071`
or `(1)/(n_(2)^(2)) = (1)/(9) - 0.071 = 0.111 - 0.071 = 0.04 or n_(2)^(2) = 25 or n_(2) = 5`