Molecular populations
Temperature plays a role in determining the relevance of electronic, vibrational, and
rotational spectral lines in room-temperature biomedical optics.
The ratio between any two state's populations is an exponential function of ∆E, the
difference in their energies, and kT, the Boltzmann energy, ratio is e-∆E/kT. Taking Ng to
be the number of molecules in the ground state (i.e., the electronic, vibrational, or
rotational quantum number is at a minimum) at body temperature (37°C), and Ne to be
the number in a particular excited state, calculate the fractional excitation Ne/(Ng + Ne)
of a two-state ensemble for the following cases:
(a) When the excited state corresponds to absorption of a 525 nm photon
hc 6.63 *10 -34 * 3.0 *10 8
E 

 3.79 *10 -19 J
9

525 *10
kT = 1.381×10−23*(37+273) = 4.2811*10-21 J
E
3.79 *10 19

 88.495
kT 4.2811 *10 21
Ne = Ng* e-∆E/kT
N g e  E / kT
Ne
e  E / kT
e -88.495



 3.69 * 10 -39
Therefore,
 E / kT
 E / kT
-88.495
N g  Ne N g  N ge
1 e
1 e
(b) When the excited state is a vibration with v (wave number) = 1030 cm-1
1
1
 
cm  0.000971cm  0.00000971m
v 1030
hc 6.63 *10 -34 * 3.0 *10 8
E 

 2.0484 *10 -20 J

0.00000971
kT = 1.381×10−23*(37+273) = 4.2811*10-21 J
E 2.0484 *10 -20

 4.785
kT 4.2811 *10 21
Ne = Ng* e-∆E/kT
N g e  E / kT
Ne
e  E / kT
e -4.785



 0.00829
Therefore,
N g  N e N g  N g e  E / kT 1  e  E / kT 1  e -4.785
(c) When the excited state is a rotation with v (wave number) = 30 cm-1
1 1
   cm  0.0333cm  0.00033m
v 30
hc 6.63 *10 -34 * 3.0 *10 8
E 

 6.02727*10-22 J

0.00033
kT = 1.381×10−23*(37+273) = 4.2811*10-21 J
E 6.02727 *10 -22

 0.14079
kT
4.2811 *10 21
Ne = Ng* e-∆E/kT
N g e  E / kT
Ne
e  E / kT
e 0.14079



 0.46486
Therefore,
N g  N e N g  N g e  E / kT 1  e  E / kT 1  e 0.14079
Based upon these calculations, is it reasonable to assume that essentially any molecule a
photon encounters in the body will be in its ground vibrational state?
Yes, it is. Because from the first calculation (a), we see that the photon on wavelength of
the order of nanometers has the ratio of the number of excited molecules to the total
number of molecules as extremely low.
It’s ground electronic state?
Yes, because from (b), the ratio is still quite small.
It’s ground rotational state? No, because from (c), the ratio is significantly large.