Here is what I did, though I see I goofed with the math, but the procedure is possibly correct if you only want a rough value. If I had the data set, I could do a more respectable job, if for some reason such a result would be useful.
Originally Posted by Nereid
Since E=hc/lambda , then the “blue” photons will have twice the energy of the far “red” ones, which are twice the wavelength. This will squash the blue end of the spectrum if we convert the curve to a photon flux curve. The peak shifts to the red. [The Sun’s shift is from 495nm (using a Planck peak @ 5850K -- the real peak is closer to 450nm, surprisingly) for a wattage curve to 695nm to a photon flux curve peak.]
As I understand it, and without an appropriate book or teacher, the integration (ie area) of the curve represents the flux density because the power stated for the y-axis is per unit wavelength. [Be sure to extrapolate into the IR band to get the bulk of the complete spectrum.] So what we want out of this spectrum is the power (ergs per second) value that is a result of the integration. The Sun, for instance, has a peak sp. irr. value of over 2,100 watts m-2 @ 1 AU, but we know the actual wattage (ie Solar Constant) is only ~ 1,361 watts m-2 @ 1 AU. So, in the Sun’s case, the value is 64% of the peak.
Arbitrarily using this same percentage would give us about 9.5 E -17 ergs s-1 cm-2 for the example galaxy spectrum.
Now choose a wavelength that best represents the place the flux distribution is balanced. Say it is about 1500nm, which has a photon energy of 1.32E-12 ergs. So dividing this energy into the power above (per cm-2) yields a photon flux rate of 7.20E-5 per cm -2. This means you would need a 1.3 meter aperture scope to get one photon per second.
Obviously, this is a very rough value given the lack of effort to get hard values, but it should be a reasonable approach to the problem.
It is much too late to study the other related posts by the more credible posters, but one concern I have regarding Vega is that it is a “blue” star (9600K) so there will be fewer photons coming from it compared to a red star, for instance, having the same apparent magnitude -- perhaps by a factor of 2 to 5, which isn't much of a variance I suppose.
I hope the others will review this approach since it is just something I did on a wing. [It was a wing that stayed on the ground for a long time, admittedly. ]
We know time flies, we just can't see its wings.