Beer-Lambet Law and Quantum Yield

[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), I_trans, 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, I_abs. 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*]
I_f = Φ_fI₀ (1-10-A)
where:
I_f = emission intensity
Φ_f = quantum yield
I₀ = incident light intensity
A = absorbance??

 

[ehild]

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), I_trans, 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, I_abs. 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?

 

[Lily Wright]

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

 

[ehild]

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.

 

[Lily Wright]

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?

 

[ehild]

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.