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: Write the energy formula. The energy of the \(n^{th}\) level of hydrogen is
\[E_n=-\frac{13.6}{n^2}\ \text{eV}.\]
Step 2: Absorption condition. When a photon is absorbed, the electron jumps from a lower level A (quantum number \(n_A\)) to a higher level B (quantum number \(n_B\)). The photon energy equals the energy difference:
\[E_{photon}=E_{n_B}-E_{n_A}=2.55\ \text{eV}.\]
So
\[-\frac{13.6}{n_B^2}-\left(-\frac{13.6}{n_A^2}\right)=2.55.\]\[13.6\left(\frac{1}{n_A^2}-\frac{1}{n_B^2}\right)=2.55.\]\[\frac{1}{n_A^2}-\frac{1}{n_B^2}=\frac{2.55}{13.6}=0.1875.\]
Step 3: Try \(n_A=2\). Then \(\dfrac{1}{n_A^2}=\dfrac{1}{4}=0.25\), so
\[\frac{1}{n_B^2}=0.25-0.1875=0.0625=\frac{1}{16}\ \Rightarrow\ n_B^2=16\ \Rightarrow\ n_B=4.\]
Step 4: Verify with actual energies. \(E_2=-\dfrac{13.6}{4}=-3.40\ \text{eV}\) and \(E_4=-\dfrac{13.6}{16}=-0.85\ \text{eV}\).
\[E_4-E_2=-0.85-(-3.40)=2.55\ \text{eV}.\ \checkmark\]
Step 5: Rule out other guesses. For \(n_A=1\): \(\frac{1}{n_B^2}=1-0.1875=0.8125\), giving \(n_B^2=1.23\), not a whole number. For \(n_A=3\): \(\frac{1}{n_B^2}=0.1111-0.1875<0\), impossible. So the only valid solution is \(n_A=2,\ n_B=4\).

\[\boxed{n_A=2\ \ (\text{state A}),\qquad n_B=4\ \ (\text{state B})}\]