Explanation

(b) Ionization energy corresponding to ionization potential
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ - 13.6\ eV$
Photon energy incident $= \ 12.1\ eV$
So,the energy of electron in excited state
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - 13.6\ + 12.1\ = - 1.5\ eV$
$ie,$ $\ \ \ \ \ \ \ \ \ E_{n}$ =$- \ \frac{13.6}{n^{2}}$ $eV$
$\ \ \ \ \ \ \ \ \ \ \ \ - 1.5\ = - \ \frac{- 13.6}{n^{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ n^{2}$ =$\frac{- 13.6}{- 1.5}\ \approx 9$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 3$
$ie,$ energy of electron in excited state corresponds to third orbit.
The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st. Thus, 3 spectral lines are emitted.