# Number of atoms trapped in an atom trap

1. Mar 17, 2012

### stigg

1. The problem statement, all variables and given/known data
a group of atoms are confined in a point like volume in a laser based atom trap, the laser light causes each atom to emit 1.0 x 10^6 photons of wavelength 780 nm every second. the sensor has area of 1 cubic centimeter and measure the light intensity emanating from the trap to be 1.6 nW when placed 25 cm away from the trapped atoms. assuming each atom emits photons with equal probability in all directions, determine the number of trapped atoms.

2. Relevant equations

3. The attempt at a solution
i honestly dont know where to begin with this problem, any guidance would be greatly appreciated, thanks

2. Mar 17, 2012

### fluidistic

Calculate the power emitted by a single atom. You have the number of photons it emits per second as well as their frequency so you can calculate the total energy emitted in 1 s, and thus get the power emitted.
The energy emitted by a single atom will be distributed homogeneously radially from it. Since your censor is 1 cm³ and inside an imaginary sphere of radius 25 cm, you can calculate how much % of the light (and hence power) coming from a single atom is received.
Can you figure out how to continue?

3. Mar 19, 2012

### stigg

i found the frequency using f=v/$\lambda$ and then the power using hf emitted by each photon then multiyply that by the number of photons released per second, is this correct?

Last edited: Mar 19, 2012
4. Mar 19, 2012

### fluidistic

I'm not sure what you mean. The "hf" term is energy, not power. With the "hf" term you get the energy of a single photon. A single atom emits 10^6 of these photons, per second.

5. Mar 19, 2012

### stigg

ah yes youre right my mistake, so i used E=hf and mulitpled it by 10^6 photons per second which would then in turn be equal to the power, correct?

Last edited: Mar 19, 2012
6. Mar 19, 2012

### fluidistic

Yes, exactly. Let P be this power. It will be distributed uniformly radially from the source. For instance, at 1 m away from the source, the power emitted by a single atom will be distributed over a surface of $4 \pi$. At a distance r, $4 \pi r ^2$. In the problem you are given a distance and the volume of the sensor. Do you have an idea how to calculate the percentage of emitted photons received in the sensor?

7. Mar 19, 2012

### stigg

first find the total intensity over the surface area of the sphere with radius 25cm and then find what fraction of that the sensor area is?

8. Mar 19, 2012

### fluidistic

Yes.

9. Mar 19, 2012

### stigg

i have the ratio of the sensor area to the total sphere area, however how do i use that to find the emmitted photons in the sensor

10. Mar 19, 2012

### fluidistic

Ok good.
This ratio is the percentage of the emitted photons that are received by the sensor, for all atoms. In the problem statement they give the information that the intensity in the sensor is 1.6 nW. You can determine the number of photons the sensor receives per second.
Let's suppose (unrealistically) that the sensor receives 100 photons per second. And that you calculated that a single atom emitts 30 photons per second. Say you calculated the ratio we're talking about as being 1%. It means that the total source (or all atoms together) emitts 100 times more photons than then ones your sensor receives. Thus, in total 10,000 photons are emitted from the source per second. There's a very small last step, let's see if you can figure it out. :)

11. Mar 19, 2012

### stigg

i divided the number of photons hitting the sensor per second by the ratio we talked about to receive the total number of photons released from the source per second i then divided this by the number emitted per atom to find the number of atoms in the source. does this sound correct?

12. Mar 19, 2012

### fluidistic

I think so :) Post your numbers just in case the result is strange.

13. Mar 19, 2012

### stigg

i calculated 1.95 x 10^12 atoms in the source, would you like my other numbers as well

14. Mar 19, 2012

### fluidistic

No need, that looks "possible" at first glance.

15. Mar 19, 2012

### stigg

sounds good, thanks a bunch for the help glad i came here it was very useful!

16. Mar 19, 2012

### fluidistic

You're welcome and feel free to use this forum (as much as I do ).