Question

Please provide legible answers so that I can understand the concept behind it and rework the problem myself.

Assume a strongly scattering, infinite medium (refractive index, n 1.35), with a chromophore, whose molar extinction coefficient is 2.5 x 105 cm 1M1, at a concentration of 0.35 HM. a) Use the microscopic Beer-Lambert law (also called the time-resolved Beer-Lambert law) to determine the ratio of the fluence rate (at a given position) to the fluence rate without the chromophore at times 0.5 ns and 1.5 ns after the emission of a short light pulse. b) This ratio can be called the attenuation factor. Why does the attenuation factor associated with the presence of the chromophore depend on the time of observation?

Answer 1

(a) As molar extinction coefficient = 2.5 x 10⁵ cm^-1M^-1

Hence for 0.35 micro Molar solution, extinction coefficient is:

2.5 x 10⁵ x 0.35 x 10^-6 cm^-1

or

8.75 x 10^-2 cm^-1

Now, speed of light in media with n = 1.35 is v = (3/1.35) x 10¹⁰ cm/s

v = 2.222 x 10¹⁰ cm/s

hence, ditance travelled between 0.5 ns to 1.5 ns is:

d = v x t = 2.222 x 10¹⁰ x 1 x 10^-9 cm

d = 22.22 cm

Now, using Beer Lamberts law:

(a) As molar extinction coefficient = 2.5 x 105 cm-1M-1 Hence for 0.35 micro Molar solution, extinction coefficient is: 2.5 x 105 x 0.35 x 10-6 cm-1 or 8.75 x 10-2 cm-1 Now, speed of light in media with n = 1.35 is v = (3/1.35) x 1010 cm/s v = 2.222 x 1010 cm/s hence, distance travelled between 0.5 ns to 1.5 ns is: d = v x t = 2.222 x 1010 x 1 x 10-9 cm d = 22.22 cm Now, using Beer Lamberts law: I = I_0 exp(- c.x) where c is extinction coefficient and x is distance travelled. hence, ratio of fluency rate at 0.5 ns Io and at 1.5 ns I is: L = 6.9884I_0/I = exp(c.x) I_0/I = exp(8.75 times 10^- 2 times 22.22) I_0/I = 6.9884 This depends upon time of observation, because light travels a finite distance within medium in a finite time

where c is extinstion coefficient and x is distance travelled.

hence, ratio of fluence rate at 0.5 ns I_o and at 1.5 ns I is :

$I_o/I = exp(8.75 \times 10^{-2}\times 22.22)$

(b) This depends upon time of observation, because light travels a finite distaance within medium in a finite time. hence, distance travelled effects the order of its attenuation as it is absorbed in the medium.