Which of the following transitions in the hydrogen atom emits a photon of the lowest frequency? (n = principal quantum number)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → n = 4 to n = 3
In a transition, a photon carries away the energy difference between the two levels, \( h\nu=E_{n_i}-E_{n_f} \). So the lowest frequency belongs to the smallest energy gap. With \( E_n=-13.6\,\mathrm{eV}/n^2 \) for hydrogen, the photon energy is:
\( \begin{aligned} h\nu &= E_{n_i}-E_{n_f} \\ &= 13.6\,\mathrm{eV}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \end{aligned} \)
For the four transitions:
\( \begin{aligned} h\nu_{2\to1} &= 13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\mathrm{eV} \\ &\approx 10.2\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} h\nu_{4\to3} &= 13.6\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\mathrm{eV} \\ &\approx 0.66\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} h\nu_{3\to1} &= 13.6\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\mathrm{eV} \\ &\approx 12.1\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} h\nu_{4\to2} &= 13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\mathrm{eV} \\ &\approx 2.55\,\mathrm{eV} \end{aligned} \)
From the above, the gap from n = 4 to n = 3 is the smallest, about 0.66 eV, because the levels crowd together as n increases. Hence, the transition n = 4 to n = 3 emits the photon of the lowest frequency.
NCERT (2026-27):
The Line Spectra of the Hydrogen Atom – “According to the third postulate of Bohr’s model, when an atom makes a transition from the higher energy state with quantum number ni to the lower energy state with quantum number nf (nf < ni), the difference of energy is carried away by a photon of frequency νif such that
\( h\nu_{if}=E_{n_i}-E_{n_f} \qquad (12.11) \)”