Respuesta :
Answer:
The concentration of an anthracene solution is [tex]c = 3.560 *10^{-5}M[/tex]
Explanation:
From the question we are told that
The incident beam [tex]P_o[/tex] is [tex]=1532[/tex]
The fluorescence intensity is [tex]I = 775[/tex]
The length of the medium is b = 0.875 cm
The molar extinction coefficient is [tex]\epsilon = 9.5 *10^3 M^{-1} cm^{-1}[/tex]
The proportionality constant k = 0.30
According to Lambert law the Absorbance of the anthracene solution is mathematically represented as
[tex]A = log (I_O/I)[/tex]
Where [tex]I_o =P_o[/tex]
and A is the Absorbance
Substituting value
[tex]A = log( (1532)/(775))[/tex]
[tex]=0.2960[/tex]
Generally beers law can be represented mathematically as
[tex]A = \epsilon c l[/tex]
where c is the concentration of an anthracene solution
Making c the subject of the formula
[tex]c = \frac{A}{cl}[/tex]
Substituting 0.875 cm for length = b ,
We have
[tex]c = \frac{0.2960}{9.5*10^{3} * 0.875}[/tex]
[tex]c = 3.560 *10^{-5}M[/tex]
The concentration of an anthracene solution is [tex]c=3.560*10^{-5} M[/tex] which can be identified using Lambert Beer's law for absorption.
Given:
The incident beam P0= 1532
The fluorescence intensity I= 775
The length of the medium is b = 0.875 cm
The molar extinction coefficient ( ϵ ) = [tex]9.5 * 10^3 M^{-1} cm^{-1}[/tex]
The proportionality constant k = 0.30
Lambert Beer's Law:
A= ϵ *l * c
Where,
A is the Absorbance
c is the concentration
l is the path length
( ϵ ) is the molar extinction coefficient
A= log (I₀ / I)
On adding values:
A= log (I₀ / I)
A= log (1532/775)
A= 0.2960
On substituting the values in the above formula we will get,
A= ϵ *l * c
c = A / ϵ *l
[tex]c= \frac{0.2960}{ 9.5* 10^3 *0.875}[/tex]
[tex]c=3.560*10^{-5} M[/tex]
Thus, the concentration an anthracene solution is [tex]c=3.560*10^{-5} M[/tex].
Find more information about Lambert Beer's law here:
brainly.com/question/26555614
