band energy modification of ferroelectric znsno 3 for photovoltaic applications b.smith, c. kons, a....
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Band Energy Modification of
Ferroelectric ZnSnO3 for photovoltaic applications
B.Smith, C. Kons, A. Datta
University of South Florida, Department of Physics
NSF REU grant # DMR-1263066 REU site in Applied Physics at USF
Florida Cluster for Advanced Smart Sensor Technologies1
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Blackbody Radiation Comps Problem
The question is broken up into 5 parts (Will discuss each individually)
Uses mainly the power per unit area over a small wavelength interval at location λ which is R(λ) radiated by an ideal thermal radiator at some temperature T.
I will refer to two constants with unchanging variables throughout, C1 and C2
2
C1 = 2πc2h = 3.747 E -16 m4 s-3 kg C2 = hc/k = 0.01440 K m
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Part A
For a Blackbody at Temperature T, find an expression for the wavelength at which the maximum power is radiated (λmax(T)). Then evaluate the expression at the temperature of the photosphere of the sun and explain it’s significance.
Took the derivative of R(λ) and set it equal to zero to find maximum and got to here:
Replace C2/λT with u and graphed to find intersection
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Intersection point occurs when u = 4.965
Final Equation =
The Sun’s photosphere is the surface of the sun and the source thermal radiation collected by solar panels that is converted into energy. 4
0 2 4 6 8 10 12
Grapical Determination of u (hc/λkT)
5ue (̂u) / e (̂u) -1
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Part B Integrate R(λ) over all wavelengths and show that the total radiated power per unit
area is proportional to T4.
By plugging in u = hc/λkT we simplify the integral to
5
Constant *
∫ hc
ukT ___( ()5 1 ___( ) eu - 1 kT
-hc( ) ___) u ___
2
1 du
Reduces to:
Constant *
∫4T ( ) hc
k ___4
eu - 1 ____ u( )
3
du
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hc k ___
Part C Find the fraction of radiated power in the region of the spectrum
below λmax(T).
We already know we can model the function with a single variable u so we find
Power radiated below λmax(T) = 0.251 or approximately 25 percent of the total area.6
u3
eu - 1______ du = ∫ u3
eu - 1______ du = ∫
infinity
0
infinity
4.965
6.4810
1.62351
Constant *
4
T ( )4
eu - 1 ____ u( )
3
du
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Solar Spectrum Curve
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Part D Starting with the expression for R(λ), derive an expression for the radiated
power per unit area as a function of frequency.
8
ν2
-c ___dλ =
dν
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Part E Plot both R(λ)dλ and R(ν)dν for the temperature of the solar photosphere. For each
R, find the value of the independent variable (λmax, νmax) at the respective R has its maximum.
9
Max = 500nmW/m^2 8.42E+13
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Solar Spectrum Curve
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Power v.s. Frequency
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Max = 3.33E+14W / m^2 1.15992E-07
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Frequency v.s. Wavelength
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Questions
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Design Slide
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R(ν)dν = C1ν3 1
c4 e(hν/kT) - 1
________ _________ dν
R’(λ)dλ = 0after
simplification
5 = Tλ (e(C2/λT) - 1)
C2e(C2/λT) _____________
λmax(T) = ukT
hc ____ = 500 nm at 5800 K
Constant *
∫ λ5
1 ___ = Constant * λ4
- 1 ____ λ = ukT
hc ____Constant * T4
=
R(λ)dλ =ukT 1
hc eu - 1____ ______ du2πc2h ( )
5 u5
eu - 1______ du = ∫
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Design Slide 2
ν
c __ λ =
ν
ν2
-c ___dλ =
dν