(d) $E = E_{1}/n^{2}$
[image:13_image31.emf]
Energy used for excitation is $12.75\ eV$
$i.e.,( - 13.6 + 12.75)eV = - 0.85\ eV$
The photons of energy $12.75\ eV$ can excite the fourth level of $H$-atom. Therefore six lines will be emitted $\left( n\frac{(n - 1)}{2}lines \right)$