Question

Calculate the flux (in [photons/second]) in a 1 [µW], 600 [nm] beam.

Answer 1

Photons of frequency \(f=600\,[\text{nm}] \)

The energy of each is \(E=hf=hc/\lambda\) \[\qquad\qquad\qquad\qquad\qquad=1240[\text{nm}\cdot\text{eV}]/600[\text {nm}]\\\qquad\qquad\qquad\qquad\qquad=2.07[\text{eV}]\]

Answer 2

So you have the energy of a single photon, so I think you take the power of the beam and divide it by the energy of the photon, since power is is watts(Joules/second).

\[1 \mu W --> 1 * 10^{-6} W\]

\[\phi = \frac{1*10^{-6}[\frac{J}{s}]*(whatever the conversion factor forJ \to eV is???)}{2.07 [\frac{eV}{photon}]}\]

I was doing some reading that said flux is usually written per unit area though, but maybe its just different terminology(?).

Answer 3

so the flux is the power per energy of the photons?

\[\Phi = P/E\\
\qquad=\frac{1\times10^{-6}[\text W]}{2.07[\text{eV}/\gamma]}\\\qquad=4.83\times10^{-7}\left[\frac{\text A\,\text V}{\text{eV}/\gamma}\right]\\\qquad=4.83\times10^{-7}[(\text C/\text s)\,(\gamma/\text e)]\times{1.602\div10^{-19}[\text C/\text e]}\\\qquad=3.016\times10^{12}[\gamma/\text s]\]

Answer 4

Yeah, the units cancel out so it becomes photons per unit time, but you can also think of it in that way.