Question

S₁ = Energy of e of H in ground state × 

= 2.25 × energy of e of H in ground state

 $\because$ One radial node

  $\therefore \mathrm{n} - \mathcal{l}$ – 1 = 1

Or  $\mathrm{n}-\mathcal{l}=2$

 $\mathcal{l}=0$

n = 2

Orbital name = 2s

For S₂ = n – $\mathcal{l}$ – 1

n – $\mathcal{l}$ = 2

n = 3, $\mathcal{l}$ = 1

Orbital = 3p ∴ $\mathcal{l}$ = 1

[ S₂ = energy of e of H in ground state ×  $\frac{3^2}{{\mathrm{n}}^2},\mathrm{n}=3]$