Solved Examples of Beer-Lambert Law
Example 1: Find the absorbance of a solution if its concentration is 1 mole/litre, the molar absorption coefficient is 6240 M/cm and the path length is 0.002 m.
Solution:
We have,
c = 1, ε = 6240 and L = 0.002
Using Beer Lambert law, we have
A = εLc
= 6240 (0.002) (1)
= 124.8
Example 2: Find the absorbance of a solution if its concentration is 1000 millimoles/litre, the molar absorption coefficient is 5342 mM/cm and the path length is 0.00001 mm.
Solution:
We have,
c = 1000, ε = 5342 and L = 0.00001
Using Beer Lambert law, we have
A = εLc
= 5342 (0.00001) (1000)
= 5.342
Example 3: Find the concentration of a solution if its absorbance is 6.85, the molar absorption coefficient is 2371 mM/cm and the path length is 321 nm.
Solution:
We have,
A = 6.85, ε = 2371 and L = 321
Using Beer Lambert law, we have
A = εLc
6.85 = 2371 (321) c
c = 6.85/761091
c = 90 nM
Example 4: Find the path length of a solution if its absorbance is 0.37, the molar absorption coefficient is 3298 mM/cm and the concentration is 75 μM.
Solution:
We have,
A = 0.37, ε = 3298 and c = 75
Using Beer Lambert law, we have
A = εLc
0.37 = (3298) (75) L
L = 0.37/247350
L = 15.23 μM
Example 5: Find the relative intensity of light absorbed by the solution if it has an absorbance of 2.
Solution:
We have, A = 2.
Using Beer Lambert law, we have
A = log10 (I0/I)
I0/I = 10 A
Putting A = 2, we have
I0/I = 20
I/I0 = 1/20
1 – I/Io = 1 – 1/20
(Io – I)/Io = 19/20
[(Io – I)/Io] × 100 = (19/20) × 100
R = 95%
Example 6: Find the relative intensity of light absorbed by the solution if it has an absorbance of 5.
Solution:
We have, A = 5.
Using Beer Lambert law, we have
A = log10 (I0/I)
I0/I = 10 A
Putting A = 5, we have
I0/I = 50
I/I0 = 1/50
1 – I/Io = 1 – 1/50
(Io – I)/Io = 49/50
[(Io – I)/Io] × 100 = (49/50) × 100
R = 98%
Example 7: Find the relative intensity of light absorbed by the solution if it has an absorbance of 1.5.
Solution:
We have, A = 1.5.
Using Beer Lambert law, we have
A = log10 (I0/I)
I0/I = 10 A
Putting A = 1.5, we have
I0/I = 15
I/I0 = 1/15
1 – I/Io = 1 – 1/15
(Io – I)/Io = 14/15
[(Io – I)/Io] × 100 = (14/15) × 100
R = 93.33%
Beer Lambert Law
Beer-Lambert Law derivation describes the relationship between the attenuation of light through a substance with the properties of that substance. It asserts that the length of a sample route and the concentration of a solution are proportional to the light’s absorbance. This law is dependent on the properties of the solution in question. It elaborates on the linear relationship between the concentration and absorbance of a solution.
Table of Content
- Beer-Lambert Law Statement
- Schematic Diagram of Beer-Lambert Law
- Beer-Lambert Law Formula
- Beer-Lambert Law Equation
- Beer-Lambert Law Derivation
- What Is Beer’s Law?
- What Is Lambert Law?
- Beer-Lambert Law Applications
- Beer-Lambert Law Limitation
- Solved Examples of Beer-Lambert Law
- Beer-Lambert law
- Define Beer-Lambert Law
- What are the limitations of Beer-Lambert Law?
- Who discovered the Beer-Lambert Law?
- Why is Beer-Lambert Law important?