Given, Energy given to hydrogen atom in ground state =∆E Let initial angular momentum be L. Then, final angular momentum be L′=L+
h
2π
Since, L=
nh
2π
. . . (i) Angular momentum of ground state (n=1) L=
1⋅h
2π
=
h
2π
Hence, from Eq. (i), L′=
h
2π
+
h
2π
=2⋅
h
2π
Final orbit level, nf=2 Now, electrons will transit from orbit n=1 to n=2 Since, ∆E=13.6Z2(
1
nf2
−
1
ni2
) . . .(ii) where, Z= atomic number, and E= transition energy of electron, while moving ni to nf and Z (hydrogen atom) =1 Substituting in Eq. (ii), we get ∆E=13.6×12(