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

6

The correct answer is Option (3) → 6

Ionization potential of H atom = 13.6 eV

Energy of photon = 12.75 eV

Hydrogen energy levels are:

$E_n = -\frac{13.6}{n^2}$ eV

Ground state: $E_1 = -13.6$ eV

Excited energy = $E_1 + 12.75 = -13.6 + 12.75 = -0.85$ eV

For $n=4$: $E_4 = -\frac{13.6}{16} = -0.85$ eV

So electron goes to $n=4$ level.

Number of possible spectral lines from an excited level $n$ is given by:

$N = \frac{n(n-1)}{2}$

For $n=4$: $N = \frac{4 \times 3}{2} = 6$

Answer: The hydrogen atom will emit 6 spectral lines.