JEE Main 31-Aug-2021 Shift 2 Solved Paper

Given that, energy of free electron,
E1=2.6‌eV
We know that, the energy of H-atom in its first excited state (n = 2),
E2=

−13.6

22

=−

13.6

4

‌eV
Now, the energy of emitted photons will be the difference ofthese two energies E1 and E2.
ΔE=E1−E2
∴hv=[2.6−(−

13.6

4

)]‌eV