A sample of 2g of hydrogen atoms are supplied with 12.75 ev/atom of energy. During deexcitation, the number of spectral lines obtained are

E₁(H)=-13.6ev/atom

Energy supplied=12.75ev/atom

energy of higher energy level= -13.6+12.75=-0.85

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}={\mathrm{E}}_1\left(\frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}\right) \\ -0.85=-13.6\left(\frac{1}{{\mathrm{n}}^2}\right) \\ \mathrm{n}=4 \end{gathered}$$

$\therefore$ no. of spectral lines formed = $\sum _{ }^{ }({\mathrm{n}}_2-{\mathrm{n}}_1) = \sum _{ }^{ }(4-1)=6$