Question

The frequency of one of the lines in Paschen series of hydrogen atom is $2.750\times {10}^{14}$ Hz. The quantum number which produces this transition is:

• A. n=2
• B. n=4
• C. l=2
• D. n=6

Answer 1

Answer: (C), (D)

n=6

For paschen series, $n_1$=3, we need to find $n_2$

v=$\frac{c}{\lambda }$ = $c\times R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

=$\frac{1}{\lambda }$ = $c\times R_H[\frac{1}{3^2}-\frac{1}{n_x^2}]$ = $n_2$ = $6$

Solve this using fractions, DO NOT substitute until the end, makes life easier that way.

The thing with the hydrogen atom is that since there is only one electron, all the orbitals - s, p, d, f would be equivalent. Hence when we are at n = 6, l can take any value, and hence l = 2 is correct as well.