The figure shows the energy level diagram of the hydrogen atom. Out of the six given transitions, which transition results in the emission of a photon of wavelength 4870 Å?: (h = \( 6.63 \times 10^{-34}\,\mathrm{J\,s} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → n = 4 to n = 2
A photon emitted in a transition carries the energy difference between the two levels, \( E = hc/\lambda \). Find the photon energy in eV and compare it with the level differences in the figure.
\( \lambda = 4870\,\mathring{\mathrm{A}} = 4.87\times10^{-7}\,\mathrm{m} \)
\( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34} \times 3\times10^{8}}{4.87\times10^{-7}} = 4.08\times10^{-19}\,\mathrm{J} \)
\( \displaystyle E = \frac{4.08\times10^{-19}}{1.6\times10^{-19}}\,\mathrm{eV} \approx 2.55\,\mathrm{eV} \)
Now the energy differences from the figure:
\( n=4 \to 2: -0.85 - (-3.4) = 2.55\,\mathrm{eV} \)
\( n=3 \to 1: -1.51 - (-13.6) = 12.09\,\mathrm{eV} \)
\( n=4 \to 1: -0.85 - (-13.6) = 12.75\,\mathrm{eV} \)
\( n=3 \to 2: -1.51 - (-3.4) = 1.89\,\mathrm{eV} \)
Only the transition from \( n = 4 \) to \( n = 2 \) gives 2.55 eV.
Shortcut: Use \( E\,(\mathrm{eV}) \approx 1240/\lambda\,(\mathrm{nm}) \): \( 1240/487 \approx 2.55\,\mathrm{eV} \), which is the 4 to 2 gap, so the other levels need not be checked in detail.