Using Bohr's postulates derive the expression for the frequency of radiation emitted when an electron in a hydrogen atom undergoes transition from a higher energy state (quantum number ni) to a lower state (nf). When an electron in a hydrogen atom jumps from energy state ni=4 to nf=3, 2, 1, identify the special series to which the emission lines belong.

In the hydrogen atom:
Radius of electron orbit, r = n²h²/4π²kme² (i)
Kinetic energy of electron: Ek = ½mv² = ke²/r
Using equation (i), we get:
Ek = ke²/4π²kme²/n²h² = 2π²k²me⁴/n²h²
Potential energy: Ep = -k(e)×(e)/r = -ke²/r
Using equation (i) we get:
Ep = -ke² × 4π²kme²/n²h² = -8π²k²me⁴/n²h²
Total energy of electron: E = 2π²k²me⁴/n²h² - 8π²k²me⁴/n²h² = -8π²k²me⁴/2n²h² = -2π²k²me⁴/n²h² = -2π²k²me⁴/h² × [1/n²]
Now, according to Bohr's frequency condition, when an electron in a hydrogen atom undergoes a transition from a higher energy state (ni) to a lower energy state (nf) is:
hv = Emi - Enf
hv = -2π²k²me⁴/h² × 1/ni² - [-2π²k²me⁴/h² × 1/nf²]
hv = 2π²k²me⁴/h² × [1/nf² - 1/ni²]
v = 2π²k²me⁴/h³ × [1/nf² - 1/ni²]
v = c2π²k²me⁴/ch³ × [1/nf² - 1/ni²]
2π²k²me⁴/ch³ = R = Rydberg constant
R = 1.097 × 10⁷ m⁻¹
Thus, v = Rc × [1/nf² - 1/ni²]
Now higher state, ni = 4
Lower state, nf = 3, 2, 1
For the transition:
ni = 4 to nf = 3, → Paschen Series
ni = 4 to nf = 2, → Balmer Series
ni = 4 to nf = 1, → Lyman Series