\(E=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10}} \mathrm{J}\)

\(=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=12.75 \mathrm{eV}\)

After absorbing a photon of energy \(12.75\,eV\), the electron will reach to third excited state of energy \(-0.85\) eV, since energy difference corresponding to \(n=1\) and \(n=4\) is \(12.75 \mathrm{eV}\)

\(\therefore\) Number of spectral lines emitted

\(=\frac{(n)(n-1)}{2}=\frac{(4)(4-1)}{2}=6\)