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 ?

The frequency of one of the lines in Paschen series of hydrogen atom i

1.	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» Solution :`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`