In the determination of the quantum yield of a photoisomerization, $\Phi,$ the following is needed

$$ \Phi = \frac{\mbox{Number of "reactions"}}{\mbox{Number of absorbed photons}} $$

The photon flux $\rho$ on the sample is the following

$$ \rho = \frac{I_0}{E} = \frac{I_0 \lambda}{hc} $$

But this is just the total number of photons hitting the sample per second per meter squared. How determine the number of photons that are absorbed?

I feel I should take into account the absorbance of the sample, but we have been told that we will measure the irradiance of the incident light, and use that to obtain the number of absorbed photons.

Is there something I am missing?


If light of intensity $I_0$ is incident on a sample or a dielectric interface, at the boundary the following equation will hold: $$I_0 = I_T + I_R + I_A$$ where $I_T$ is the transmitted light, $I_R$ is the reflected light and $I_A$ is the fraction of light which is absorbed by be medium. The absorption is usually calculated from the Beer–Lambert law $$\phi_T = \phi_i e^{-\tau} $$ where $\phi_T$ is the transmitted flux, $\phi_i$ the incident flux and $\tau$ the absorption coefficient, which for an inhomogeneous sample (thickness $t$) with $N$ types of absorbing species is given by $$\tau = \sum_{i=0}^{N}\sigma_i \int_0^t n_i(x)dx$$ where $\sigma_i$ is the absorption cross-section and $n_i$ is the number density of each species. Now in your case, if the only information provided is the irradiance of the sample, then I would think that the assumption is that all the photons not reflected by the sample are absorbed i.e. $\tau\rightarrow\infty$ and the material is essentially opaque and $$I_0 = I_R +I_A$$

Hope this helps :)

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