Question

Energy of the \(n^{th}\) state of a hydrogen atom is \(E_n=-\dfrac{13.6}{n^2}\ \text{eV}\). A hydrogen sample is prepared in a particular excited state A. Photons of energy \(2.55\ \text{eV}\) get absorbed by the sample to take some electrons to a particular further excited state B. Find the quantum numbers of states A and B.

Hint:

Set the photon energy equal to \(E_{n_B}-E_{n_A}\) and find whole-number levels; \(2.55\ \text{eV}\) matches the jump from \(n=2\) to \(n=4\).

Answer 1

Step 1: Tabulate the low-lying hydrogen levels. Using \(E_n=-13.6/n^2\) eV:
\(E_1=-13.60\) eV, \(E_2=-3.40\) eV, \(E_3=-1.51\) eV, \(E_4=-0.85\) eV, \(E_5=-0.544\) eV.

Step 2: Search for a \(2.55\) eV gap. Absorption of \(2.55\) eV means the difference between the final and initial level equals \(2.55\) eV. Scan the pairs:
\(E_2\to E_3:\ 1.89\) eV; \(E_2\to E_4:\ 2.55\) eV; \(E_1\to E_2:\ 10.2\) eV; \(E_3\to E_4:\ 0.66\) eV.
Only the \(2\to4\) transition matches exactly.

Step 3: Confirm. \(E_4-E_2=(-0.85)-(-3.40)=2.55\) eV, so the sample was initially in state A with \(n=2\) and the absorbed photon lifts electrons to state B with \(n=4\).

Step 4: Consistency note. Since the sample is described as 'a particular excited state', A cannot be the ground state (\(n=1\)); \(n=2\) is indeed an excited state, and \(n=4\) is a further excited state, matching the wording.

\[\boxed{\text{State A}:n=2,\qquad \text{State B}:n=4}\]