Let's say I have a radio tower with frequency $v[HZ]$ and power output of $P[W]$ radiating uniformly across all directions, and I need to find the number of photons crossing an area of $1m^2$ per $1sec$ a distance $1m$ away.

This is my solution which apparently is wrong :

I'll find the intensity per area by taking the total power output $P$ and dividing it by the area of the sphere created that is $1m$ away, which is $4\pi$.

So we get $I = \frac{P}{4\pi} [\frac{J}{m^2s}] $

This is the energy per unit area per second, I shall divide the energy by the energy of a single photon to get the number of photons per unit area:

$E = hv[J] $,

$\frac{I}{E} = \frac{P}{4\pi E} [\frac{photons}{m^2s}] $ is the photons per unit area

I'll multiply by the area which is $1m^2$ and get $\frac{P}{4\pi E} [\frac{photons}{s}]$ photons are passing the area per second.

What is wrong with my solution and how can I do it right ?

  • $\begingroup$ what's the answer you're looking for? $\endgroup$ – JEB Aug 9 '18 at 1:09

Is your detector "area" a flat surface?

Your logic is sound for the case where your detector is a flat surface, far away from the source, and the normal of the detector is pointing towards the source.

If your detector area is just a section of the sphere with an area of $1 \mathrm{m^2}$ then your calculation is still right. I think the only way you can be wrong is if your detector is a flat surface. Then, since it's so close to the source, different parts of your detector are at different distances to the source and see a different intensity per area.

  • $\begingroup$ I mean it's a question from homework where I got a different answer than the solutions, so I needed to know my logic is fine and I'm not missing an integral part. Thanks for the answer. $\endgroup$ – user3575645 Aug 9 '18 at 8:25
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    $\begingroup$ What did the solutions say? $\endgroup$ – Alex Aug 9 '18 at 13:10
  • $\begingroup$ They used the numbers for power : $P = 200kW$, for frequency: $v = 101.1MHz$ and it gave them the answer $6.39*10^{21}$ photons per second, but if I enter these values in the formula I wrote for photons per second I get the value of $1.782*10^{29}$, a totally different magnitude. The precise wording they used after giving those values is : "find the number of photons crossing a unit area in unit time 1 mile away from the the radio station". Their answer is in $ft^2$ and mine is in $m^2$ I don't know how they calculated that maybe its the weird units they decided to use that changes it? $\endgroup$ – user3575645 Aug 9 '18 at 17:25
  • $\begingroup$ You might be confusing yourself by using the symbol $m$ sometimes for miles and sometimes for meters. $\endgroup$ – Alex Aug 9 '18 at 17:31
  • $\begingroup$ I used $m$ only for meters, since they said "unit area.." etc I used meters squared for area, seconds for time, and 1 meter for distance, as it wasn't clear for me what units they expect me to use $\endgroup$ – user3575645 Aug 9 '18 at 22:46

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