# converting power spectrum to photon flux density

So I'm having trouble converting units and was hoping somebody could point out where I've gone wrong... It seems I'm missing something fundamental.

a Power Spectrum has units $kW/m^2-\mu m$ for the y-axis and microns for the x-axis. I take it, for the $y$ axis that this actually means it has units $kW m^{-3} \mu m^{-1}$ instead of $kW m^{-3} \mu m$.

a Photon Flux Density has units $10^{17} cm^{-2} s^{-1} eV^{-1}$ for the $y$ axis and photon energy ($eV$) for the x-axis.

Here is my attempt to convert them... We know $W=\frac{J}{s}$, so we can write $$10^3 J s^{-1} m^{-2} \mu m^{-1}$$ and we know how to convert $m$ to $cm$, so $$10^3 J s^{-1} 10^{-4} cm^{-2} \mu m^{-1}$$ I interpret this to mean: so many joules hitting a certain area every second at a specific wavelength. Now we want to speak about the number of photons hitting a given area every second for a particular area and for a given photon energy.

So we could multiply by the wavelength $\lambda$ ($\mu m$) to cancel out the extra reciprocal microns and we'd get, $$10^3 J s^{-1} 10^{-4} cm^{-2} \lambda$$ Now I know the energy of a photon can be given by $E=\frac{hc}{\lambda}$ and $J$ are units of energy, so $$10^3 \frac{hc}{\lambda (\mu m)} s^{-1} 10^{-4} cm^{-2} \lambda$$ and $hc =1.24 eV \mu m$ so, $$10^3 \frac{1.24 eV \mu m}{\lambda (\mu m)} s^{-1} 10^{-4} cm^{-2} \lambda$$ which I get to mean,

$$10^3 1.24 eV s^{-1} 10^{-4} cm^{-2}$$ $$10^{-1} 1.24 eV s^{-1} cm^{-2}$$ which is almost what I wanted, except I'm missing a $10^{18}$ and my eV is not reciprocal. Can somebody help get me get this?!

Alternatively I tried multiplying by $\frac{\lambda}{hc}$ (although I cannot justify why) and then we get, $$10^{-1} J s^{-1} cm^{-2} \mu m^{-1} \frac{\lambda}{hc}$$

Then I let $hc = 1.99 10^{-25} J m$ and we get, $$10^{-1} J s^{-1} cm^{-2} \mu m^{-1} \frac{\lambda (\mu m^{-1}}{1.99 10^{-25} J m}$$ Simplify because the joules cancel out and the microns,

$$10^{24} s^{-1} cm^{-2} \frac{\lambda}{1.99 m}$$

Now after simplifying I'm left with $$\frac{\lambda 10^{24}}{1.99} s^{-1} cm^{-2} m^{-1}$$

The internet tells me $m^{-1} = 8.1 10^5 eV$ and after plugging in I get $$\frac{8.1 \lambda 10^{19}}{1.99} s^{-1} cm^{-2} eV$$

which is much closer but again the $eV$ are killing me. Help, please!

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Let's start from the beginning. On the one hand, you can get the power per unit area between two wavelengths $\lambda_1$ and $\lambda_2$ by integrating power per unit area per unit wavelength against wavelength: $$\frac{P}{A} = \int_{\lambda_1}^{\lambda_2} \underbrace{F_\lambda}_{(\mathrm{kW}/\mathrm{m}^2)/\mu\mathrm{m}} \ \underbrace{\mathrm{d}\lambda}_{\mu\mathrm{m}}.$$ Now what rate $R$ of photons would you see in this band, per unit area? The answer is you have to divide the integrand by the energy per photon before integrating: $$\frac{R}{A} = \int_{\lambda_1}^{\lambda_2} \frac{F_\lambda}{E} \ \mathrm{d}\lambda. \tag{1}$$

On the other hand, you have another way of thinking about the rate. In this case you can bin based on energy (in $\mathrm{eV}$) rather than wavelength. I'll call the measure of counts per unit area per unit time per energy bin by the symbol $G_E$ in analogy with $F_\lambda$.1 Then $$\frac{R}{A} = \int_{E_2}^{E_1} \underbrace{G_E}_{(1/(\mathrm{cm}^2\cdot\mathrm{s}))/\mathrm{eV}} \ \underbrace{\mathrm{d}E}_{\mathrm{eV}}.$$ Note that $E_2 < E_1$ if $\lambda_1 < \lambda_2$. Before I can compare this to (1), though, it helps to have the integration variables be of the same unit. A little change of variables tells us $$\frac{R}{A} = \int_{\lambda_1}^{\lambda_2} G_E \left(-\frac{\mathrm{d}E}{\mathrm{d}\lambda}\right) \ \mathrm{d}\lambda. \tag{2}$$

Comparing (1) and (2), you should be convinced $$F_\lambda = -G_E E \frac{\mathrm{d}E}{\mathrm{d}\lambda}.$$ Given the energy of a photon is $E = hc/\lambda$, this clearly tells us $$F_\lambda = G_E \frac{(hc)^2}{\lambda^3}. \tag{3}$$

Now (3) is ultimately all you need. Whatever units you use for these quantities is up to you. However, in the interest of completeness, I'll show the conversion factors for the units being used in this particular case. The three relations we need are

• $hc = 1.24~\mathrm{eV}\cdot\mu\mathrm{m}$,
• $1~\mathrm{kW} = 6.24\times10^{21}~\mathrm{eV}/\mathrm{s}$, and
• $1~\mathrm{m} = 100~\mathrm{cm}$.

Putting everything together yields $$F_\lambda = G_E \frac{(1.24~\mathrm{eV}\cdot\mu\mathrm{m})^2}{\lambda^3} \left(\frac{1~\mathrm{kW}}{6.24\times10^{21}~\mathrm{eV}/\mathrm{s}}\right) \left(\frac{100~\mathrm{cm}}{1~\mathrm{m}}\right)^2,$$ or $$\underbrace{F_\lambda}_{(\mathrm{kW}/\mathrm{m}^2)/\mu\mathrm{m}} = 2.46\times10^{-18} \frac{\overbrace{G_E}^{(1/(\mathrm{cm}^2\cdot\mathrm{s}))/\mathrm{eV}}}{{\underbrace{\lambda}_{\mu\mathrm{m}}}^3}.$$

1$G_E$ is my completely nonstandard notation, while $F_\lambda$ is actually used in the literature.

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This is a much better answer! –  Sankaran Jan 20 '13 at 7:59
This is fantastic! Thank you!! –  sciencenewbie Jan 21 '13 at 17:44

It is not clear to me why you are converting one to the other. Here is my understanding, see if it helps:

The power spectrum is like a histogram. The x axis is wavelength and the y axis has $\mu m^{-1}$. But what it means is for any 1 micron slice/bin you can read off the power density. so Essentially, lets say I want to know how much power per unit area is coming to me as red photons I read the y axis directly at that wavelength. I don't multiply by wavelength. This is how solar spectrum is given (google AM1.5 spectrum)

Next the photon density says for $eV$ of energy I have $10^{17}$ photons per unit area and time for any color.

Therefore I can use both these things to calculate how many red photon I receive: If power/area corresponding the red wavelength is $x kW/m^2$. Then the number of red photons is $10^{17}* x*10^3/10^4*1/q$ where $q$ is the electronic charge.

I have not really encountered photon flux density before (and some googling was not very helpful) so if I have interpreted that wrong, then this doesn't make much sense. But I am positive about the first part that you don't need to multiply by wavelength in the power spectrum

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I want to convert one to the other because a book does it without justification and makes it sound trivial, but it is non-trivial to me until I'm able to follow them. I am taking the AM 1.5 spectrum and converting it to these units so I can then integrate and obtain a photocurrent by multiplying by $q$ and $A$ (area). –  sciencenewbie Jan 19 '13 at 20:10