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Chapter 4: Problem 68

At noon on a clear day, sunlight reaches the earth's surface at Madison, Wisconsin, with an average power of approximately \(1.0 \mathrm{~kJ} \cdot \mathrm{s}^{-1} \cdot \mathrm{m}^{-2}\). If the sunlight consists of photons with an average wavelength of \(510 \mathrm{~nm}\), how many photons strike a \(1.0-\mathrm{cm}^{2}\) area per second?

Short Answer

Expert verified
Approximately \(2.57 \times 10^{15}\) photons hit a \(1.0 \text{ cm}^2\) area per second.

Step by step solution

01

Convert the power to watts

The incoming power per square meter is given as \(1.0 \text{ kJ} \cdot \text{s}^{-1} \cdot \text{m}^{-2}\). Convert this to watts: \(1.0 \text{ kJ} \cdot \text{s}^{-1} = 1000 \text{ W}\).
02

Calculate the energy per photon

Use the formula for energy of a photon: \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \text{ Js}\) is Planck's constant, \(c = 3.00 \times 10^8 \text{ m/s}\) is the speed of light, and \(\lambda = 510 \text{ nm} = 510 \times 10^{-9} \text{ m}\) is the wavelength. Compute \(E = \frac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^8 \text{ m/s})}{510 \times 10^{-9} \text{ m}}\).
03

Set up to find photon count per square meter

We know the surface receives \(1000 \text{ W/m}^2\), which is \(1000 \text{ J/s/m}^2\). Determine the number of photons per second striking a square meter by using the formula: Number of photons = Total energy per second / Energy per photon.
04

Adjust for 1 cm虏 area

1 cm虏 is \(1/10000\) of a m虏, so to find the number of photons hitting a \(1.0\text{ cm}^2\) area, divide the number of photons per second per square meter by 10,000.
05

Calculate and conclude

Substitute the values and calculate: The energy per photon is approximately \(E \approx 3.89 \times 10^{-19} \text{ J}\). Therefore, the approximate number of photons striking a \(1.0 \text{ cm}^2\) area per second is \( \frac{1000}{3.89 \times 10^{-19}} / 10000 \approx 2.57 \times 10^{15}\).

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Key Concepts

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

Energy Per Photon
Energy per photon is a crucial concept in understanding how light behaves at a microscopic level. Photons are tiny packets of energy, and each one carries energy that depends on its wavelength. The shorter the wavelength, the higher the energy per photon.
\[ E = \frac{hc}{\lambda} \]
This formula calculates the energy \(E\) of a single photon:
  • \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34}\) Joule-seconds (Js).
  • \(c\) is the speed of light, about \(3.00 \times 10^8\) meters per second (m/s).
  • \(\lambda\) is the wavelength of the photon, which is often given in nanometers (nm).
Using the energy formula, we can determine how much energy a photon carries, which is essential in calculations involving solar energy absorption.
Solar Power
Solar power measures the amount of energy from sunlight hitting the Earth's surface. It is crucial for understanding the energy input from the sun, which can be harnessed for various applications like electricity generation. The power received on Earth can be described in terms of watts per square meter (W/m虏).
For example, at noon on a sunny day in Madison, Wisconsin, the average solar power is about \(1000 \text{ W/m}^2\). This tells us that each square meter of the Earth's surface receives an energy flow of 1000 Joules every second from sunlight. Knowing the solar power helps us understand the potential energy available for conversion into electricity.
Wavelength
The wavelength is a defining feature of a photon and directly affects the energy it carries. Light behaves as both waves and particles, and wavelength is a measure of the distance between successive peaks of these waves. In terms of photons:
  • Longer wavelengths correspond to less energy per photon.
  • Shorter wavelengths carry more energy per photon.
Measure wavelength in meters (m) but often use smaller units like nanometers (nm) for convenience, especially when discussing light. For instance, a wavelength of 510 nm describes visible light within the green range of the spectrum. Understanding wavelength helps us track the distribution of energy across different frequencies of light.
Conversion Of Units
Unit conversion is a foundational skill in solving physics problems, especially those involving energy and light. Converting units involves changing measurements from one unit to another while maintaining the same quantity. This skill is vital for:
  • Accurately calculating energy, such as converting kilojoules to joules.
  • Expressing wavelength from nanometers to meters for formula compatibility.
  • Translating area measurements from square centimeters to square meters.
For example, converting the power from \(1.0 \text{kJ} \cdot \text{s}^{-1}\) (kilojoules per second) to watts (W) is essential for accurate solar power analysis since 1 kJ/s equals 1000 W. Such conversions ensure consistency across all parts of a calculation.

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Most popular questions from this chapter

The de Broglie wavelength of electrons used in a particular electron microscope is \(96.0 \mathrm{pm}\). What is the speed of one of these electrons?

In order to resolve an object in the electron microscope, the wavelength of the electrons must be close to the diameter of the object. What kinetic energy must the electrons have in order to resolve a DNA molecule, which is \(2.00 \mathrm{~nm}\) in diameter? Take the mass of an electron to be \(9.11 \times 10^{-31} \mathrm{~kg}\).

The first four energy levels in arbitrary units of energy of an unknown element are listed below: \(\begin{array}{lrrrrrr}n & 1 & 2 & 3 & 4 & \ldots & \infty \\ E & -20 & -9 & -4 & -3 & \ldots & 0\end{array}\) A gaseous sample of the element is bombarded by photons with energies in these units of \(20,16,15,11\), 9, and \(1 .\) Which of these photons could potentially be absorbed by an electron in the \(n=1\) state of the element? In each case indicate the transition that would occur.

A campus radio station broadcasts at \(89.9 \mathrm{MHz}\) (megahertz) on the FM dial. What is the wavelength of this transmission in meters?

The photocells that are used in automatic door openers are an application of the photoelectric effect. A beam of light strikes a metal surface, from which electrons are emitted, producing an electric current. When the beam of light is blocked by a person walking through the beam, the electric circuit is broken, thereby opening the door. If the source of light is a solid-state gallium-arsenide diode laser that emits light at a wavelength of \(840 \mathrm{~nm}\), would copper be a satisfactory metal to use in the photocell? The threshold frequency of copper is \(1.01 \times 10^{15} \mathrm{~Hz}\).
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