BSc Nursing GNM 2019 Solved Paper

Ionization energy corresponding to ionisation potential =−13.6eV
Incident photon energy =12.1eV
So, the energy of electron in excited state =−13.6+12.1=−1.5eV
i.e. ‌‌En=−‌

13.6

n2

eV
⇒‌‌−1.5=‌

−13.6

n2

⇒‌‌n2=‌

−13.6

−1.5

≈9
∴‌‌n=3
i.e. energy of electron in excited state corresponds to third orbit. The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3 rd to 1 st and 2 nd to 1 st. Thus, 3 spectral lines are emitted.