# Beer-Lambet Law and Quantum Yield

• Lily Wright
In summary, a cell containing gaseous IBr is irradiated by a continuous wave laser, resulting in the production of excited Br atoms. The energy of the photons is 3.73 x 10-19 J, equivalent to 2.328 eV or 18777.26 cm-1. The power rating of the laser is 5 W, resulting in 1.34 x 1019 photons per second in the laser beam. Using the Beer-Lambert Law, the intensity of the transmitted radiation is found to be 4.495 x 1015 photons per second. The quantum yield of Br* can be expressed as Φ = k[Br*]/Iabs, and with a quantum
Lily Wright

## Homework Statement

A cell of length 10 cm, containing gaseous IBr at 300 K and 5 Pa pressure is irradiated by a continuous wave laser operating at a wavelength of 532 nm and a power of 5 W.
(i) Calculate the energy of the photons. Express your answer in units of J, cm-1 and eV. From the power rating of the laser, how many photons per second are contained in the laser beam?
(ii) If the absorption coefficient of IBr at 532 nm is σ = 400 dm3 mol-1 cm-1 , calculate the intensity of the transmitted radiation (i.e. that which passes through the cell), Itrans, using the Beer-Lambert Law
(iii) Upon irradiation, IBr undergoes photodissociation to produce Br atoms in an excited state, Br* . Give an expression for the quantum yield, Φ, of Br* in terms of the rate of production k[Br* ] and the intensity of absorbed radiation, Iabs. If the quantum yield is 0.1, use the result from (ii) to calculate the rate of formation of Br*

## Homework Equations

where I0 is the is the incident radiation intensity calculated in (i) C is the concentration of IBr in the cell and L the length of the cell

## The Attempt at a Solution

(i) E = hc/λ = 3.73 x 10-19 J (=2.328 eV; 18777.26 cm-1)
5W = 5 J s-1
5/3.73 x 10-19 = 1.34 x 1019 photons per second
(ii) is I0 the same as the value for photons per second here? I don't know the concentration of IBr in the cell?
(iii) completely lost here, found these equations: d[Br*] dt = -kf[A*] - knr[A*] = -(kf + knr)[A*]
If = ΦfI0 (1-10-A)
where:
If = emission intensity
Φf = quantum yield
I0 = incident light intensity
A = absorbance??

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Lily Wright said:

## Homework Statement

A cell of length 10 cm, containing gaseous IBr at 300 K and 5 Pa pressure is irradiated by a continuous wave laser operating at a wavelength of 532 nm and a power of 5 W.
(i) Calculate the energy of the photons. Express your answer in units of J, cm-1 and eV. From the power rating of the laser, how many photons per second are contained in the laser beam?
(ii) If the absorption coefficient of IBr at 532 nm is σ = 400 dm3 mol-1 cm-1 , calculate the intensity of the transmitted radiation (i.e. that which passes through the cell), Itrans, using the Beer-Lambert Law
(iii) Upon irradiation, IBr undergoes photodissociation to produce Br atoms in an excited state, Br* . Give an expression for the quantum yield, Φ, of Br* in terms of the rate of production k[Br* ] and the intensity of absorbed radiation, Iabs. If the quantum yield is 0.1, use the result from (ii) to calculate the rate of formation of Br*

## Homework Equations

View attachment 79430
where I0 is the is the incident radiation intensity calculated in (i) C is the concentration of IBr in the cell and L the length of the cell

## The Attempt at a Solution

(i) E = hc/λ = 3.73 x 10-19 J (=2.328 eV; 18777.26 cm-1)
5W = 5 J s-1
5/3.73 x 10-19 = 1.34 x 1019 photons per second
(ii) is I0 the same as the value for photons per second here? I don't know the concentration of IBr in the cell?

The laser beam has some cross section D, and I 0 the intensity of the incident beam is energy/cross section / seconds. You can consider the intensity as the number of incident photons per second , and and you need the concentration of the IBr in the volume irradiated by the laser in the cell of length L=0.1 m.
The pressure and temperature of the gas in the cell are given. How do you get the number of moles in the volume illuminated by the laser and the concentration of IBr?

Thanks! OK sorry but how do I find the cross-section? and then I would just divide number of protons per second by this?
Using pV=nRT I can rearrange for number of moles n = pV/RT so if I have the volume I can work out no of moles I just don't really see where I'm supposed to get that from

You know the length of the cell, and the relevant volume is that which is irradiated by the laser beam. So V= L*D. But you need the concentration, which is n/V=p/(RT), so do not worry about the volume.
The formula for the transmitted intensity ##I_{trans}=I_0 e^{-\sigma c L }## includes the concentration in moles/ dm3 and the length of the cell in cm. σ = 400 dm3 mol-1 cm-1.

Ohhh ok thanks. So I0 is 1.34 x 1019, the number of photons per second? So we have 1.34 x 1019 e(-400x2x10-3x10) which gives me 4.495 x 1015, and this is also in photons per second?

You have to use the proper units. The absorption coefficient is given as σ = 400 dm3 mol-1 cm-1. You got the concentration in mol/m3, transform it to mol/dm3.
The transmitted and incident intensities can be given in photons/s, yes.

## 1. What is the Beer-Lambert Law?

The Beer-Lambert Law is a mathematical equation that describes the relationship between the concentration of a substance in a solution and the amount of light that is absorbed by that solution. It is commonly used to measure the concentration of a compound in a solution, such as in the case of spectrophotometry.

## 2. How is the Beer-Lambert Law used in chemistry?

In chemistry, the Beer-Lambert Law is used to determine the concentration of a substance in a solution by measuring the absorbance of light passing through the solution. This is often done using a spectrophotometer, which measures the amount of light absorbed by a sample at various wavelengths.

## 3. What is the relationship between the Beer-Lambert Law and the quantum yield?

The Beer-Lambert Law and the quantum yield are both used to measure the concentration of a substance in a solution. However, the Beer-Lambert Law is based on the amount of light absorbed by the solution, while the quantum yield is a measure of the efficiency of a substance in emitting light. These two concepts are related, as the quantum yield can be used to calculate the absorbance of a solution according to the Beer-Lambert Law.

## 4. What factors can affect the accuracy of the Beer-Lambert Law?

Several factors can affect the accuracy of the Beer-Lambert Law, including the purity of the solution, the path length of the light through the solution, and the concentration of the compound being measured. In addition, the wavelength of light used and the solvent used can also impact the accuracy of the measurement.

## 5. What are the limitations of the Beer-Lambert Law?

The Beer-Lambert Law is based on several assumptions, including the absence of scattering and the linearity of the absorbance with concentration. These assumptions may not hold true in all cases, and therefore, there may be limitations to the accuracy of the law. Additionally, the Beer-Lambert Law is only applicable to solutions with low to moderate concentrations, and may not be accurate for highly concentrated solutions or those with strong light-absorbing substances.

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