IIT JEE Advanced 2010 Paper 2

For S1 (spherically symmetrical), number of radial node = 1.
Therefore, n - l - 1 = 1.
⇒ n = 2 (as l = 0 for spherically symmetrical s orbital)
For S2 state, we have
ES2 = - 13.6 ×

Z2

n2

Also, given that EH in ground state = –13.6
Therefore,
ES2 = - 13.6 ×

9

n2

= - 13.6 ⇒ n = 3
Since the number of radial node = 1, substituting in n - l - 1 = 1, we get l = 1.
So, the state is S1 is 2s and S2 is 3p.