Ionization potential of a hydrogen atom is 13.6 V. Hydrogen atom in the ground state is excited by monochromatic radiation of photon energy 12.75 eV. The number of spectral lines emitted by hydrogen atom will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6
An ionization potential of 13.6 V means the ground-state energy is \( -13.6\,\mathrm{eV} \). The photon raises the atom to the level whose energy is 12.75 eV higher.
\( \displaystyle E = -13.6\,\mathrm{eV} + 12.75\,\mathrm{eV} = -0.85\,\mathrm{eV} \)
\( \displaystyle E_n = -\frac{13.6}{n^{2}}\,\mathrm{eV} \;\Rightarrow\; n^{2} = \frac{13.6}{0.85} = 16 \;\Rightarrow\; n = 4 \)
From \( n = 4 \) the electron can return to the ground state by every possible route, and each pair of levels gives one line. The number of spectral lines is
\( \displaystyle N = \frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6 \)
\( 4 \rightarrow 3,\; 4 \rightarrow 2,\; 4 \rightarrow 1 \)
\( 3 \rightarrow 2,\; 3 \rightarrow 1,\; 2 \rightarrow 1 \)