Hydrogen atom in ground state is excited by a monochromatic radiation of λ= 975 A°. Number of spectral lines in the resulting spectrum emit

Energy of the photon, $E=\frac{h c}{\lambda}$

$E=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10}} \mathrm{J} \\ =\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}\\ =12.75 \mathrm{eV}\\$

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 eV.

Number of spectral lines emitted = $\frac{n(n-1)}{2} = \frac{4(4-1)}{2}=6$

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