Given that, ionisation potential of hydrogen atom, E1=−13.6‌eV Energy supplied = Energy absorbed by hydrogen atom. ΔE=12.1‌eV i.e. final energy of excited state, E2=
−13.6‌eV
n2
We know that, E2−E1=ΔE⇒E2=ΔE+E1 Substituting the value, we get −
13.6
n2
=[12.1+(−13.6)] ⇒−
13.6
n2
=−1.50⇒x2=9.07⇒n≃3 Hence, by absorbing 12.1 eV energy electron jump into 2nd excited state (n=3). Then, number of spectral lines emitted by H-atom =