In a Hydrogen atom there is transition of electron from one orbit to another. Choose the correct option given below in which the emitted photon will have the highest frequency?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → n = 2 to n = 1
The frequency of the emitted photon is proportional to the energy difference between the two levels, \( h\nu = E_i - E_f \), with \( E_n = -13.6/n^{2}\,\mathrm{eV} \) for hydrogen.
\( \displaystyle h\nu = 13.6\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right)\,\mathrm{eV} \)
\( \displaystyle 3 \to 2:\ 13.6\left(\frac{1}{4} - \frac{1}{9}\right) \approx 1.89\,\mathrm{eV} \)
\( \displaystyle 6 \to 2:\ 13.6\left(\frac{1}{4} - \frac{1}{36}\right) \approx 3.02\,\mathrm{eV} \)
\( \displaystyle 2 \to 1:\ 13.6\left(1 - \frac{1}{4}\right) = 10.2\,\mathrm{eV} \)
\( \displaystyle 4 \to 2:\ 13.6\left(\frac{1}{4} - \frac{1}{16}\right) = 2.55\,\mathrm{eV} \)
The energy gaps shrink rapidly at higher levels, so the jump to the ground state \( n = 1 \) (Lyman series) gives the largest energy, and hence the highest frequency.