In a hydrogen atom, a transition of an electron takes place from \( n = 3 \) to \( n = 2 \) orbit. The energy of the emitted photon will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.89 \) eV
The energy of level n of hydrogen is \( E_n = -13.6/n^{2} \) eV, and the photon carries away the difference of the two levels.
\( \displaystyle \Delta E = 13.6\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = 13.6\times\frac{5}{36} \approx 1.89\,\mathrm{eV} \)
This is the first line of the Balmer series (red light, about 656 nm).