2. Factors influencing intensity of spectral lines
• Amount of sample
• Beer-Lambert Law
• Population of energy states
• Boltzmann Distribution ∆𝑱 = + 𝟏 − 𝟏
• Spectroscopic selection rules
• Allowed and forbidden
J=0
J=1
3. Population of Energy states
• The probability of a transition taking place is the most important factor
influencing the intensity of an observed rotational line.
• This probability is proportional to the population of the initial state involved in
the transition.
• The population of a rotational state depends on two factors.
• Factor 1.
• The number of molecules in an excited state with quantum number J, relative to
the number of molecules in the ground state , NJ/N0 is given by
the Boltzman distribution
4. Population of Energy states
• Boltzmann distribution and degeneracy.
•
𝑵𝒇
𝑵𝒊
=
𝒈𝒇
𝒈𝒊
𝒆𝒙𝒑(−
∆𝑬
𝑲𝑻 = 𝒆𝒙𝒑
−𝑩𝒉𝒄 𝑱(𝑱+𝟏)
𝒌𝑻
• Nf and Ni are the population of molecules in energy
levels ef and ei with degeneracy gf and gI.
• where k is the Boltzmann constant and T the absolute
temperature. This factor decreases as J increases.
5. Intensity of spectral lines
• Factor 2
• The second factor is the degeneracy of the rotational state, which is equal to 2J+1.
This factor increases as J increases. Degeneracy is existence of two or more energy
states which have exactly same energy
• From quantum mechanics, the degeneracy of each eJ level = 2J + 1
• For j=0 non degenerate
• For j=I 3 fold degenerate
• For j=2 5fold degenerate
• The total relative population at energy 𝐸𝐽
• 𝑬𝑱 ∝ (2J+1)𝒆𝒙𝒑
−(𝑬𝑱
𝑲𝑻)
7. • Like the angular momentum of an electron,
the angular momentum of a rotating
molecule is quantized and must have specific
spatial orientations depending on the
quantum number "J". In fact, there can be
only 2J+1 spatial orientations for each
rotational levels. This 2J+1 is the
"degeneracy.“
• As a result of these restriction, rotational
energy levels occur only at discrete values
Degeneracy
8. Energy and Angular Momentum
• E=
1
2
I𝜔2………1
• P =I𝜔…………..2
• I is moment of inertia and 𝜔 is angular velocity, P is angular momentum
• 2E= I𝜔2……….3
• 2EI=𝐼2 𝜔2………4 From eq. 2 and 4
• P =√ 2EI…………5
• Energy can be written as
• E = J(J+1).
ℎ2
𝐼.8𝜋2……….6 so 2EI= J(J+1).
ℎ2
4𝜋2…………..7 substituting values from equation 7 in
equation 5
• P =√ J(J+1).
ℎ2
4𝜋2 or P =√ J(J+1) units
9. Applications
• J=1
• P =√ J(J+1)units = √2=1.41
• P vector has length 1.41 units
• Degeneracy
• 2J+1 =3 Three degenerate axis at J=-1,0,+1
• J=2
• -2,-1,0,+1, +2
• =2(2+1) = √ 6 units