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Question

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 spectral series to which the emission lines belong.

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Solution

In the hydrogen atom,
Radius of electron orbit,
r=n2h24π2kme2 ..... (i)
Kinetic energy of electron,
Ek=12mv2=ke22r
Using equation (i), we get
Ek=ke224π2kme2n2h2=2π2k2me4n2h2
Potential energy,
Ep=k(e)×(e)r=ke2r
Using equation (i), we get
Ep=ke2×4π2k kme2n2h2=4π2k2me4n2h2
Total energy of electron,
E=2π2k2me4n2h24π2k2me4n2h2
=2π2k2me4n2h2
=2π2k2me4h2×(1n2)
Now, according to Bohr's frequency condition when electron in hydrogen atom undergoes transition from higher energy state to the lower energy state (nf) is,
hv=EniEnf
or \left(\frac{-2 \pi^2k^2me^4}{h2} \times \frac{1}{n^2_f} \right)\)
or hv=2π2k2me4h2×(1n2f1n2i)
v=2π2k2me4h3×(1n2f1n2i)
v=c2π2k2me4ch3×(1n2f1n2i)
2π2k2me4ch3 = R (Rydberg constant)
R=1.097×107 m1
Thus,
v=Rc×(1n2f1n2i)
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


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