# Photons on a cell surface

1. Dec 30, 2012

### dscot

1. The problem statement, all variables and given/known data

Hi,

So I have this problem regarding a photovoltaic cell I found part 1 easy and part 2 wasn't actually too bad either but I couldn't finish the calculation. I managed to get to the stage where I have an integral , the problem is that the upper limit is infinity and I'm not really sure how to handle that, the question also mentions that the problem cannot be solved analytically?

The full question - http://screencast.com/t/C96Kqs3M [Broken]
solution: http://screencast.com/t/szPlkSEAYE6A [Broken]

2. Relevant equations

Provided in solution link ( see above)

3. The attempt at a solution

So I found a table of standard integrals that matched the stage I got up to and I used the value there and carried on solving as normal but I ended up with the wrong final answer. The value I found for the integral was 2.4 but I think this is the wrong method as it only valid for 0 to infinity?

Last edited by a moderator: May 6, 2017
2. Dec 30, 2012

### Dick

Yes, 2.4 is for lower limit 0 and your lower limit is substantially different from 0. What is the correct lower limit? Do you have a tool like Mathematica you can use to evaluate the correct number? And integrating to get a number flux won't really help you get the total power. You want to integrate energy*dJ, don't you?

Last edited by a moderator: May 6, 2017
3. Jan 2, 2013

### dscot

Hi Dick,

Sorry some of the solution got cut off, the full version is here: http://screencast.com/t/Ss2toUdI [Broken]

I calculate the actual value of the lower limit to be, 1.6911*10^-19 if I have performed the calculation correctly.

Sadly, I think we may need to do something like this in an exam so I won't be able to use any of the tools you mentioned.

Thanks again
Dave

Last edited by a moderator: May 6, 2017
4. Jan 2, 2013

### Dick

If the lower limit were as small as 10^(-19) you could safely ignore the difference between that and 0. But I get the lower limit to come out about 2.5. You can't really ignore that.

Last edited by a moderator: May 6, 2017
5. Jan 5, 2013

### dscot

Hi Dick,

I'm still unsure about this so I may have to speak to my teacher again.

In the meantime any ideas on how to do the next part? When inputting the values into the Q equation from my last post I get 1.10794× 10^-26 W which is wrong? We have the value for N,h and f so it should simply be a case of plugging them into the equation?

6. Jan 5, 2013

### Dick

You've got the right N even though the solution left off a 1/c^2 factor. Then sure, you should just multiply that by hf_G. I get the stated answer. What numbers did you use? For a shortcut hf_G is just 1.1 eV, right?

7. Jan 5, 2013

### dscot

Hi Dick,

So I'm using N (8.62*10^25) from the answer at the moment, h = 6.63^10-34 and I converted 1.1 ev into joules so 1.76 × 10-19. But those values gave me an incorrect value for Q. The answers show it should be 15.2MWm-2 but my answer was much smaller?

If I understand you correctly the correct method is to multiply 8.62*10^25(N) by 1.1? This also gives me the wrong answer so I think I might have misunderstood you?

8. Jan 5, 2013

### Dick

hf_G is the band gap energy of 1.1 eV. If you want to actually find f_G then it's (1.1 eV)/h=2.66*10^14/s. The way I think you are doing it you are multiplying by h twice instead of cancelling it out. Yes, N*(1.1 eV) is correct. Convert to that joules.

Last edited: Jan 5, 2013
9. Jan 5, 2013

### dscot

Wow thank you so much! I finally get it now :)

I really appreciate you being so patient with me , I know I'm really slow at learning things :D

Hopefully I'll be in a position to help others someday just like you have helped me!

Thanks again!

10. Jan 5, 2013

### Dick

No problem. If you want to go back the integral issue then try and compute the lower bound again. It's just plugging in numbers. You'll probably get it right now. Then you can ask Wolfram Alpha to evaluate it for you. It's the easiest way I can think of to do numerical integration.

Last edited: Jan 5, 2013