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Chapter 13: Problem 12

Which photons are most responsible for heating up a silicon photovoltaic panel in full sun: blue photons or infrared photons (beyond \(1.1 \mu \mathrm{m}\) )?

Short Answer

Expert verified
Answer: Infrared photons (beyond \(1.1 \mu \mathrm{m}\)) are most responsible for heating up a silicon photovoltaic panel in full sun.

Step by step solution

01

Understand the given wavelengths and their corresponding energies.

The problem mentions blue photons (visible light) and infrared photons (beyond \(1.1 \mu \mathrm{m}\)). The energy of a photon can be calculated using the equation: \(E = \dfrac{hc}{\lambda}\), where E is the energy, h is Planck's constant (\(6.626 × 10^{-34} \ J\cdot s\)), c is the speed of light (\(2.998 × 10^8 \ m/s\)), and \(\lambda\) is the wavelength of the photon. Blue photons fall in the wavelength range of roughly 450-495 nm, or \(4.5 \times 10^{-7} \ m\) to \(4.95 \times 10^{-7} \ m\). Infrared photons beyond \(1.1 \mu \mathrm{m}\) have wavelengths larger than \(1.1 \times 10^{-6} \ m\).
02

Calculate the energy range of blue and infrared photons.

Using the given wavelength ranges, we can determine the energy ranges for both blue and infrared photons using the energy equation E = hc/λ. For blue photons: \(E_\text{blue} = \dfrac{(6.626 × 10^{-34} \ J\cdot s)(2.998 × 10^8 \ m/s)}{4.5 \times 10^{-7} \ m} \text{ to } \dfrac{(6.626 × 10^{-34} \ J\cdot s)(2.998 × 10^8 \ m/s)}{4.95 \times 10^{-7} \ m}\) For infrared photons beyond \(1.1 \mu \mathrm{m}\): \(E_\text{infrared} \leq \dfrac{(6.626 × 10^{-34} \ J\cdot s)(2.998 × 10^8 \ m/s)}{1.1 \times 10^{-6} \ m}\)
03

Compare the energies of blue and infrared photons.

We can see that blue photons have higher energies than infrared photons as their wavelengths are smaller. When these photons strike the silicon photovoltaic panel, they will transfer their energies to the electrons in the silicon material of the panel.
04

Determine the heating effect on the photovoltaic panel.

The silicon photovoltaic panel can only absorb photons with energies above its bandgap energy, which is approximately 1.1 eV. Blue photons have sufficient energy to excite electrons and contribute to the panel's electricity production. However, lower energy infrared photons will not be able to excite electrons across the bandgap but will still transfer their energy to the panel's material, causing a heating effect.
05

Conclusion:

Infrared photons (beyond \(1.1 \mu \mathrm{m}\)) are most responsible for heating up a silicon photovoltaic panel in full sun. This is because they have lower energies than blue photons and their energy is not sufficient to contribute to electricity production but still contributes to the heating effect on the panel.

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Most popular questions from this chapter

If typical insolation is \(200 \mathrm{~W} / \mathrm{m}^{2}\), how much land area would be needed for a \(15 \%\) efficient flat PV array supplying an average of \(10,000 \mathrm{~W}\) for every person in the U.S. (population 330 million). If arranged in a square, how large is the side-length of this array? Draw it on top of a state of your choice, to scale.

Many people have an instinctive reaction to discount the \(<20 \%\) PV panel efficiency as disappointingly low-perhaps thinking they should hold out for higher. Present a multi-point argument about why the efficiency is actually pretty good, and why in practice it is plenty good enough to be practical.

The outcome of Problem 5 indicates that a hot light bulb filament emits thousands of times more power per unit area than human skin. Yet both a human and a light bulb may emit a similar amount of light \(^{105}\) - both around \(100 \mathrm{~W}\). Explain how both things can be true?

Figure \(13.7\) shows operational curves of a PV cell for different levels of illumination. If the illumination is low and the panel continues to operate at maximum power, \(^{110}\) which changes the most compared to full-sun operation: the voltage or the current? Why might lower light (fewer photons) directly connect to a lower current based on the physics of PV operation?

A 30 year study by the National Renewable Energy Lab \(^{112}\) indicates that in San Diego, a typical year delivers an annual average of \(5.0 \mathrm{kWh} / \mathrm{m}^{2} /\) day of insolation for a flat panel facing straight up. Convert this to \(\mathrm{W} / \mathrm{m}^{2}\).
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