Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

According to the table in Problem 26, San Diego can expect an annual average solar yield of \(5.7 \mathrm{kWh} / \mathrm{m}^{2} /\) day when the panel is tilted to the site latitude and facing south. \(^{116}\) If a household seeks to produce a modest \(8 \mathrm{kWh}\) per day using \(16 \%\) efficient panels, how large will the array need to be? Express as an area in square meters, and in side length for a square of the same area.

Human bodies also glow by the same physics as the sun or a light bulb filament, only it is too far out in the infrared for the human eye to see. For familiar objects (and human skin) all in the neighborhood of \(300 \mathrm{~K}\), what is the approximate wavelength of peak blackbody radiation, in microns?

If aiming for a particular power output \(^{109}\) from a PV array, describe explicitly/quantitatively how PV panel efficiency interacts with the physical size (area) of the array. For instance, what happens if the efficiency doubles or is cut in half, while keeping the same target output?
See all solutions

Recommended explanations on Environmental Science Textbooks

Ecology Research

Read Explanation

Ecological Conservation

Read Explanation

Pollution

Read Explanation

Environmental Policy

Read Explanation

Energy Resources

Read Explanation

Living Environment

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free