Question

A line is in the Paschen series of the emission spectrum of atomic hydrogen is observed at a wave number of 7800 cm^-1. Deduce the upper state principal quantum number for this transition.

Answer 1

for paschen series , final state is n=3

wavelength = 1/7800 cm

= 1.282*10^-4 cm

= 1282 nm

Use: Rydberg formula:

1/lambda = R*(1/nf^2 - 1/ni^2)

1/(1282*10^-9) = 1.097*10^7 * (1/3^2 - 1/ni^2)

(1/3^2 - 1/ni^2) = 0.071106

1/ni^2 = 0.0400

ni^2 = 25

ni = 5

Answer: 5