Solar Panel - Voltage and Current from Spectral Radiance

[Hannah Vioula]

Homework Statement

i)A beam of light at wavelength 600nm, with spectral width of 4nm and spectral radiance of 1300 Wm^-2μm_-1, illuminates an area of 10cm² . What is the voltage this beam could generate if converted to energy?
ii) What current could this beam produce?
iii) How would the current change if the width of the beam were 5 meV?

Homework Equations

i) Voltage = hf / charge = change in energy / charge = h*c / q*λ

ii)
Spectral Radiance= e_λ = hf φ_λ
Where, Spectral Flux Density = φ_λ
f=c/λ
Current Density = q * φ_λ * δλ = I / A

iii) Absolutely no idea what to use.
Maybe, E (in eV)=1.24 / λ (in μm)

The Attempt at a Solution

[/B]
i)
I need max voltage available, so I ignore spectral width.
Voltage = h*c/ (600*10^-9 * q)= 2.060 Volts

ii)
I get flux density first
Spectral radiance= e_λ
Spectral flux density = φ_λ
Rearranging equation -- -- > Spectral Radiance= e_λ = hf φ_λ
φ_λ = e_λ * λ / h*c = 1300 * (600*10^-9) / h*c = 4.58*10²¹
Rearranging equation, Current Density = q * φ_λ * δλ = I / A
I = q * φ_λ * δλ * A = 3.68 mA

iii) Not a the faintest of idea of what to do here...

 

[haruspex]

spectral radiance of 1300 Wm^-2μm_-1

I'm puzzled by this statement. Spectral radiance is for the case where light is coming from a surface at some distance. To get the total power we have to consider the area of the emitting surface as well as the solid angle the receiving surface subtends at each point of the emitting surface. I'm guessing this should say spectral irradiance. See e.g. https://en.wikipedia.org/wiki/Radiance.
I assume you meant 1300 Wm^-2μm^-1

Energy of wavelength 602nm = h*c / (602nm) = 3.31*10-19
Energy of wavelength at 598nm = h*c / (598nm) = 3.28*10-19

No point in working those two out separately, just work with the average here.
That's in Joules, right?

Voltage = Change in energy / charge = -0.014V

That's not what I get, but I could be wrong. Please post your working.

 

[Hannah Vioula]

Thanks, I re edited my original post and I definatly have part 1 and part 2 correct as I checked with my study group who all confirmed they have that answer.

I just don't know where to get part iii

 

[haruspex]

Thanks, I re edited my original post and I definatly have part 1 and part 2 correct as I checked with my study group who all confirmed they have that answer.

I just don't know where to get part iii

Given the units, I have to guess that beam width refers to the range of wavelengths. In your original post you calculated the Joules per photon at each end of the 4nm range. The difference would be the beam width in this sense. What was that beam width in units of meV?