As we know
\(E_n = \frac{-2\pi^2 Me^4 z^2}{n^2h^2(4\pi\epsilon_o)^2}\)
For hydrogen
∴ E1 (ground state) = \(\frac{-2\pi^2M_He^4(1)^2}{h^2(4\pi\epsilon_o)^2 (1)^2} = E_H\)
For deuterium
E2 (First excited state) = \(\frac{-2\pi^2M_Oe^4(1)^2}{h^2(4\pi\epsilon_o)^2(2)^2}\)
\(= \frac{-2\pi^2M_Oe^4}{h^2(4\pi\epsilon_0)^24}\)
∵ MO > MH
∴ Magnitude of E2 is greater than \(\frac{E_H}{4}\) but there is a negative sign.
∴ E2 is smaller than \(\frac{E_H}{4}\).