# I Calculating how many Joules are required to make a photon

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1. Nov 8, 2017

### reese houseknecht

A double covalent bond of nitrogen (N2) is 15.58eV Now lets go into a example. if i have a laser that is 477nm and i do 1240/477nm then i get 2.6eV per photon.

Now if i do (15.58eV / 2.6eV) it equals ~6 photons.

Now to figure out how many Joules are required for 1 photon I do
(6.626*10^-34S * 3.00*10^17nm/s) / 477nm which equals 4.167e-19 J, and for 6 photons thats 2.5002e-18 J.

Now if i want to get 6 photons in a pulse then I would need 2.5002e-18 J right? Now that is in Joules i need per nanosecond, which is 2.5002e-9 watts. All i want to know is why does it require such low amounts of energy to do this, when all i see is that you need ALOT of watts to ionize. Now as i was typing this i thought that maybe that equation shows how many joules of energy are in each photon, but yet this video
says "Joules required per photon". But if this equation is not right then what equation should i use.

(also this is my first time on this forum so if anything is wrong or im not in the right place let me know thank you)

2. Nov 8, 2017

### kuruman

Can you explain why you do "1240/477nm"? Where did this come from?

On edit: Can you also explain what exactly you are trying to calculate? Your title suggests that you want to know how to calculate the energy of a photon in Joules. It seems you already know how to do that.

3. Nov 8, 2017

### mjc123

To be pedantic, the bond in N2 is a triple bond. But leaving that aside:
Do you really only want to ionise one molecule of N2 per pulse? No wonder the power is low. You need high power to produce and maintain a significant concentration of ionised molecules.

4. Nov 8, 2017

### reese houseknecht

So 1240 is some equation i found online, and it seems to work quite well, so your saying i can ionize an atom with that low energy but i probably wont see anything

5. Nov 8, 2017

### mjc123

1240 nm is the wavelength of a 1 eV photon.
What are you actually trying to do?

6. Nov 8, 2017

### reese houseknecht

I think 1240 is h*c and then divided by wavelngth is the ev in each photon, im trying to understand hiw photoionization works. Right now im learning about the multiphoton process. I currently dont know why the wnergy required is so low. But i see now that that is to ionize 1 atom and you probably wouldnt see anylight so i need to recalculate to account for 100000s of atons

7. Nov 8, 2017

### Khashishi

Looking at your first post, you seem to be VERY confused.
The energy of a photon is $E = h \nu$ or $E = hc/\lambda$ where $h$ is Planck's constant, $\nu$ is the frequency, $c$ is the speed of light in vacuum and $\lambda$ is the wavelength.
You can easily calculate the energy for any given wavelength.
Atoms are small, so it shouldn't surprise you that it doesn't much of a joule to ionize an atom, since a joule is a macroscopic scale quantity.

8. Nov 8, 2017

### Staff: Mentor

You won't split a nitrogen bond with multiple 2.6 eV photons. It is technically not impossible but you need a ridiculous power density to make the probability of 6 photons interacting at the same time relevant. But if we ignore that:
2.5*10-18 J = 15.6 eV. Right.
Where does the nanosecond come from?

9. Nov 8, 2017

### reese houseknecht

ok i just found out new information, i am using a 150ns pules laser. so i just read that you need ridiculous amounts of watts to be focused because all the photons wont hit the atom at once, now even if i had a peak pulse intensity of 1,666,666,666 watts in each 150ns pulse, what is my probability of ionizing nitrogen in the air. (also im not trying to split the bond i just need an electron to move to its outer most energy level and fall back in creating light)

10. Nov 8, 2017

### Staff: Mentor

Do you want to ionize it or do you want to excite it? These are different processes.

Forget processes that involve multiple photons at the same time with your laser. If your energy per photon is not sufficient to directly induce a transition in nitrogen, something you might still get is ionization from heating - the laser heats a small spot in the air enough to make a plasma there.

11. Nov 8, 2017

### reese houseknecht

I mean excite (photoionize) i know that ionize is removing the electron completely. So your saying screw the whole exact process thing. If i have enough energy to cause great amounts of heat focused on a small area i will see a plasma dot?

12. Nov 8, 2017

### Staff: Mentor

Yes.

13. Nov 9, 2017

### reese houseknecht

so how many watts do i need, how can i calculate how many i will need to see a plasma dot? also am i targeting specific atoms or just any that are in the air, i say this because if im putting billions of watts in a pulse 150ns does it really matter what atom is focused upon or will it just make plasme dot no matter what?

14. Nov 9, 2017

### Staff: Mentor

You can't target specific atoms, the focus will always be much larger than the typical distance between atoms (and you can't track individual atoms with your setup anyway). The power will depend on various parameters.

What do you actually want to do?

15. Nov 10, 2017

### reese houseknecht

I want to make a plasma dot to be very very vague. But i dont just want someone to tell me what exact laser i need to do it and thats that. I want to understand the process. So i understood photoionization and multiphoton ionization, now i found out that you dont have to be super specific. Just by focusing mass amounts of energy per pulse (watts) to a point and you will make plasma. Judging by my description of what i think happens you can tell i dont know to much about this piece of ionization or if your still ionizing at that point. Also what i want to know if electronvolts still matters in doing this?

16. Nov 10, 2017

### Staff: Mentor

You can check existing systems.

Here is an example - a 100 GW peak power laser with 35 fs pulses (3.5mJ/pulse).
Another example - the second system has a power of only 200 MW.

17. Nov 10, 2017

### reese houseknecht

thank you for the post! see we dont want to do femtoseconds, we want to do nanoseconds. it says you can its just more dangerous. we plan on using anywhere from 5mw to 250mw laser at an adjustable frequency of 1hz-1000khz, based on there statistics then this should work. if i find the right values right?

18. Nov 11, 2017

### Staff: Mentor

mW or MW?

200 MW with a suitable focus work as you can see. I'm not sure about lower values.

Even at 1 Hz, a 1 ns pulse at 200 MW leads to an average power of 0.2 W, which is quite a lot. At 1000 kHz you have an average power of 200 kW, which is completely unreasonable.
There is a reason these systems use fs pulses.

19. Nov 11, 2017

### reese houseknecht

milliwatts not mega

20. Nov 11, 2017

### reese houseknecht

Im just gonna put here what i just emailed a professor so maybe you can help me out, I am a advanced programmer trying to build a project requiring lasers, so you can see my lack of expertise in the area. However, I have been doing a lot of research on how to photoionize air, such as using the eV of N2 in the air (15.58) and then calculating the wavelength i need to ionize that which is 80nm. That is the single photon process and is deadly because of its x-ray requirements. then i found the multiphoton process which was way better in terms of readily available lasers. The problem is i just dont understand it fully. Now i found out that if you focus massive amounts of watts down to a point you can achieve making a "plasma dot".
So i started with a simple equation W/Hz = J per second. so 0.25W/1Hz = 0.25J per second, which is 250mJ per second. I then do mJ/Pulse duration = Watts per pulse. So i then plugged in my values 250mJ/150e-9S = 1,666,666,666 Watts. This number is insanely huge and does'nt make sense why it would be this high. Is my equation right or am i messing something up? Also how many watts do i need to focus down to a point to achieve this "dot of plasma". I look forward to your response.

Last edited: Nov 11, 2017