Question

Let the ground state electronic energy of hydrogen atom be E_H eV. The first excited state energy of deuterium atom (in eV) is:

(1) E_H / 4  

(2) E_H / 2

(3) higher than E_H / 4

(4) lower than E_H / 4

[Ans: (4)]     

Why is the answer (4) and why not (1)? Somebody, plz explain.

Answer 1

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) = \(\frac{-2\pi^2M_Oe^4(1)^2}{h^2(4\pi\epsilon_o)^2(2)^2}\) 

\(= \frac{-2\pi^2M_Oe^4}{h^2(4\pi\epsilon_0)^24}\)

∵ M_O > M_H

∴ Magnitude of E₂ is greater than \(\frac{E_H}{4}\) but there is a negative sign.

∴  E₂ is smaller than \(\frac{E_H}{4}\).