$E=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10}} \mathrm{J}$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=12.75 \mathrm{eV}$
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
$=\frac{(n)(n-1)}{2}=\frac{(4)(4-1)}{2}=6$
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