AP EAPCET 23-Aug-2021 Shift 2 Paper

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 =

n(n−1)

2

=

3(3−1)

2

=3