Overhead sunlight arrives on the surface of the earth at an intensity of about \(1,000 \mathrm{~W} / \mathrm{m}^{2}\). How many photons per second strike a solar panel whose area is \(1.6\) square meters, if the typical wavelength is \(\lambda=0.5 \mu \mathrm{m} ?\)

Short Answer

Expert verified
Answer: Approximately 4.02 × 10^21 photons per second.

Step by step solution

01

Calculate energy from intensity and area

To find the energy, multiply the intensity of sunlight (1,000 W/m^2) by the area of the solar panel (1.6 m^2): Energy (E) = Intensity (I) × Area (A) E = 1,000 W/m^2 × 1.6 m^2 E = 1,600 W
02

Calculate frequency from wavelength

Dividing the speed of light (c ≈ 3 × 10^8 m/s) by the wavelength (λ = 0.5 × 10^-6 m), we get the frequency (v): Frequency (v) = Speed of light (c) / Wavelength (λ) v = (3 × 10^8 m/s) / (0.5 × 10^-6 m) v ≈ 6 × 10^14 Hz
03

Calculate energy of a photon

Next, we use Planck's constant (h ≈ 6.63 × 10^-34 J·s) to calculate the energy of a photon: Energy of a photon (E_photon) = Planck's constant (h) × Frequency (v) E_photon = (6.63 × 10^-34 J·s) × (6 × 10^14 Hz) E_photon ≈ 3.98 × 10^-19 J
04

Calculate the number of photons per second

Finally, divide the energy (E = 1,600 W) by the energy of a photon (E_photon ≈ 3.98 × 10^-19 J) to find the number of photons per second (N): Number of photons per second (N) = Energy (E) / Energy of a photon (E_photon) N = (1,600 W) / (3.98 × 10^-19 J) N ≈ 4.02 × 10^21 photons/s The solar panel receives approximately 4.02 × 10^21 photons per second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photon Energy Calculation
The calculation of a photon's energy is essential in understanding how solar panels convert light into electricity. Photons are particles of light, and each one carries a certain amount of energy determined by its wavelength or frequency. To calculate this energy, we use the fundamental relationship derived from quantum physics, where the energy (\( E \)) of a photon is the product of Planck's constant (\( h \text{, which is approximately }6.63 \times 10^{-34} \text{ J·s}\text{, and} \text{the frequency} \text{(} \text{v} \text{)} \text{of the light} \).

The formula to find a photon's energy is:
\[ E_{\text{photon}} = h \times u \]
Given the frequency, which we can calculate by dividing the speed of light (\( c \text{, roughly }3 \times 10^8 \text{ m/s}\text{)} \text{by the wavelength} \text{(} \lambda \text{)} \), we find the energy of an individual photon. This understanding enables us to quantify the amount of energy a solar panel can harness from sunlight, leading to applications in solar energy technologies and the calculation of the efficiency of photovoltaic cells.

For the given exercise, knowing the intensity of sunlight and the area of the solar panel allows us to calculate the number of photons striking the panel per second, which in turn gives us insight into the potential electrical energy that can be generated.
Planck's Constant Application
Planck's constant (\( h \text{)} \) is a fundamental constant in quantum mechanics that plays a pivotal role in the energy of photons. Its application extends far beyond a simple number; it represents the quantization of energy and is essential in the field of quantum physics. Within the context of solar energy, Planck's constant is used to calculate the energy of photons, which can be absorbed by solar panels to generate electricity.

When applied together with the frequency of the light, Planck's constant gives us precise knowledge about the energy a photon carries. This understanding is crucial since solar panels are designed to capture this energy and convert it into a usable electrical form. By computing the energy of each photon and knowing the total number of photons that hit a panel, we can estimate the maximum energy output that a solar panel might produce under ideal conditions.

The application of Planck's constant in our textbook exercise allowed us to identify the energy each photon possesses, bridging the gap between the abstract quantum world and practical solar energy harvesting.
Solar Radiation Intensity
Solar radiation intensity refers to the power per unit area received from the Sun in the form of electromagnetic radiation. Measured in watts per square meter (\( W/m^2 \text{)} \), it is a crucial factor in calculating the potential energy available from sunlight for solar panels. The intensity of solar radiation can vary based on factors such as the time of day, atmospheric conditions, and geographical location.

In the context of the textbook exercise, an intensity of roughly 1,000 W/m^2 is considered, which is close to the solar constant, the average amount of solar radiation that reaches the Earth's upper atmosphere on a surface perpendicular to the Sun's rays. By understanding solar radiation intensity, we can gauge the maximum energy that a given area of a solar panel can receive and thus predict the potential electricity generation.

Additionally, solar radiation intensity informs us about the concentration of photons arriving at the solar panel, and when combined with the knowledge of photon energy, it contributes to understanding the overall efficiency of solar panels. This concept is fundamental for anyone interested in solar energy adoption and maximizing the utilization of this renewable energy source.

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Most popular questions from this chapter

The study from Problem 24 finds that a flat panel facing south and tilted at various angles \(^{113}\) relative to the horizontal produce the following annual average yields in units of \(\mathrm{kWh} / \mathrm{m}^{2} /\) day: What tilt delivers the best yield for the year, and how much better is this (in percent) than a flat plate facing straight up? What tilt appears to result in minimal seasonal variation?

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.

What are two reasons that blue photons are disadvantaged in terms of having their energy contribute to useful current in silicon photovoltaics?

One way to look at solar payback time time is to note that an installed system will cost something like \(\$ 3,000\) for each \(\mathrm{kW}_{\mathrm{p}}\) (peak capacity), and that you'll produce \(x \mathrm{kWh}\) from that \(1 \mathrm{~kW}_{\mathrm{p}}\) array if your region gets \(x\) hours of full-sun- equivalent on average. Since each kWh of electricity costs something like \(\$ 0.15\), it becomes straightforward to compute the value per day as \(\$ 0.15 x\), and determine how long to match the \$3k investment. The result is independent of the actual array size, depending only on the cost per \(W_{p}\), the solar yield at your location, and the cost of electricity. What would the payback time be, in years, if the cost is \(\$ 3 / W_{p}, 17\) the yield is 6 hours per day of full-sun-equivalent, and electricity in your region costs \(\$ 0.15 / \mathrm{kWh}\) ?

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?

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