A hydrogen atom in its ground state is irradiated by light of wavelength 970A°. Taking hc/e=1.237×10−6eV m and the ground state energy of hydrogen atom as −13.6eV, the number of lines present in the emission spectrum is

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A hydrogen atom in its ground state is irradiated by light of wavelength $970 A °$ . Taking $h c / e = 1.237 \times 10^{- 6} e V$ m and the ground state energy of hydrogen atom as $- 13.6 e V$ , the number of lines present in the emission spectrum is

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answer is 6.

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Detailed Solution

Energy of radiation incident $= \frac{h c}{λ}$ = $\frac{12370}{970}$ =12.75eV
Energy of Ground state $e^{-}$ after absorbing 12.75 eV photon is 13.6-12.75=0.85 eV
This belongs to 4th state
Number of lines in the emission
Spectrum is $\frac{n (n - 1)}{2}$ =6