Question 6.4: The first four energy levels for the hydrogen atom are shown......
The first four energy levels for the hydrogen atom are shown in the diagram on the next page. The lowest energy state is labeled E_1, and the higher states are labeled E_2, E_3, and E_4, respectively. When a hydrogen atom undergoes a transition from E_3 to E_1, it emits a photon with λ = 102.6 nm. Similarly, if the atom undergoes a transition from E_3 to E_2, it emits a photon with λ = 656.3 nm. Find the wavelength of light emitted by an atom making a transition from E_2 to E_1.
Strategy The energy-level diagram shows us the relationship between the various energy levels involved. The vertical direction in the diagram is energy. So, if we represent each transition by an arrow, as shown, then the length of that arrow will be proportional to the energy lost by the atom in that transition. From the diagram, we can see that the lengths of the arrows for the E_3 to E_2 transition and for the E_2 to E_1 transition must add up to the length of the arrow for the E_3 to E_1 transition. (Another way of thinking about this is that an atom going from the E_3 state to the E_1 state must lose the same amount of energy whether or not it stops at E_2 along the way.) Because we know the wavelengths for two of the transitions, we can find the corresponding photon energies. That will let us find the energy for the third transition, and then finally we can convert that to wavelength.
We first convert the given wavelengths to energies. For the E_3 → E_1 transition:
E_{\mathrm{3}\rightarrow1}={\frac{h c}{\lambda}}={\frac{(6.626\times10^{-34}\,\mathrm{J}\,s)(2.998\times10^{8}\,\mathrm{m}\,s^{-1})}{102.6\,\mathrm{nm}}}\times{\frac{10^{9}\,\mathrm{nm}}{1\,\mathrm{m}}}=1.936\times10^{-18}\,\mathrm{J}
For the E_3 → E_2 transition:
E_{3\rightarrow2}={\frac{h c}{\lambda}}={\frac{(6.626\times10^{-34}\,{\mathrm{J}}\,s)(2.998\times10^{8}\,{\mathrm{m}}\,s^{-1})}{656.3\,{\mathrm{nm}}}}\times{\frac{10^{9}\,{\mathrm{nm}}}{1\,{\mathrm{m}}}}=3.027\times10^{-19}{\mathrm{~J}}
From the diagram, we can see that
E_{3→1} = E_{3→2} + E_{2→1}
So,
E_{2\to1}=E_{3\to1}-E_{3\to2}=1.936\times10^{-18}\mathrm{\,J}-3.027\times10^{-19}\mathrm{\,J}=1.63\times10^{-18}\mathrm{\,J}
Now we just need to convert from energy to wavelength to reach the desired answer.
\lambda_{2\rightarrow1}={\frac{h c}{E_{2\rightarrow1}}}={\frac{(6.626\times10^{-34}\,{\mathrm{J}}\,s)(2.998\times10^{8}\,{\mathrm{ms}}^{-1})}{1.633\,{\mathrm{nm}}\times10^{-18}{\mathrm{~J}}}}=1.216\times10^{-7}\,{\mathrm{m}}
Because we know that wavelengths are most often given in nanometers, we might choose to convert that to 121.6 nm.
Analyze Your Answer Our result lies between the two wavelengths we were originally given, which is consistent with the spacing of the energy levels in the diagram.
Check Your Understanding A hydrogen atom undergoing a transition from the E_4 level to the E_3 level emits a photon with a wavelength of 1.873 μm. Find the wavelength of light emitted by an atom making the transition from E_4 to E_1.