For emission of energy, electron of H-atom must fall from higher energy state to lower energy state. Hence, options (a) and (c) are not possible. In option (b), n1=2,n2=1 ∴ Energy of emitted photon is given as E′=−13.6Z2(‌
1
n12
−
1
n22
) ⇒hv′‌=−13.6Z2(
1
22
−‌
1
12
) v′=
13.6Z2
h
×
3
4
...(i) In option (c), n1=2,n2=6, hence energy of emitted photon is given as E′′=−13.6Z2(
1
62
−
1
22
) ⇒hv′′=13.6Z2×
8
36
⇒v′′=
13.6Z2
h
×
8
36
...(ii) Hence, from Eqs. (i) and (ii), we get v′>v′′