/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 The Solar Constant measured by E... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Solar Constant measured by Earth satellites is roughly \(1400 . \mathrm{W} / \mathrm{m}^{2}\). Though the Sun emits light of different wavelengths, the peak of the wavelength spectrum is at \(500 . \mathrm{nm}\) a) Find the corresponding photon frequency. b) Find the corresponding photon energy. c) Find the number flux of photons arriving at Earth, assuming that all light emitted by the Sun has the same peak wavelength.

Short Answer

Expert verified
Question: Calculate the photon frequency, energy, and number flux of photons arriving at Earth if the Solar Constant is 1400 W/m² and the peak wavelength emitted by the Sun is 500 nm. Solution: a) The corresponding photon frequency is \(6.00 x 10^{14} Hz\). b) The corresponding photon energy is \(3.98 x 10^{-19} J\). c) The number flux of photons arriving at Earth is \(3.52 x 10^{21} \frac{photons}{m^2\cdot s}\).

Step by step solution

01

Find the Corresponding Photon Frequency

To find the frequency of the light at the peak wavelength, we can use the equation \(v = \frac{c}{\lambda}\). We know the peak wavelength is 500 nm (\(5.00 x 10^{-7} m\)) and the speed of light is \(3.00 x 10^8 \frac{m}{s}\). \(v = \frac{3.00 x 10^8 \frac{m}{s}}{5.00 x 10^{-7} m}\) \(v = 6.00 x 10^{14} Hz\) The corresponding photon frequency is \(6.00 x 10^{14} Hz\).
02

Find the Corresponding Photon Energy

To find the energy of the photons with the frequency we've calculated, we can use the equation \(E = h \cdot v\). We know the photon frequency is \(6.00 x 10^{14} Hz\) and Planck's constant is \(6.63 x 10^{-34} Js\). \(E = (6.63 x 10^{-34} Js)(6.00 x 10^{14} Hz)\) \(E = 3.98 x 10^{-19} J\) The corresponding photon energy is \(3.98 x 10^{-19} J\).
03

Find the Number Flux of Photons Arriving at Earth

To find the number flux of photons arriving at Earth, we can use the equation \(\text{Flux} = \frac{\text{Solar Constant}}{E}\). We know the Solar Constant is 1400 \(\frac{W}{m^2}\) and the photon energy is \(3.98 x 10^{-19} J\). Flux = \(\frac{1400 \frac{W}{m^2}}{3.98 x 10^{-19} J}\) Flux = \(3.52 x 10^{21} \frac{photons}{m^2\cdot s}\) The number flux of photons arriving at Earth is \(3.52 x 10^{21} \frac{photons}{m^2\cdot s}\).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Photon Frequency
Understanding photon frequency is essential because it determines the color and energy of light. The frequency of a photon refers to the number of times the wave peaks pass a point in one second, measured in hertz (Hz). The higher the frequency, the more energy the photon has. In our exercise, the frequency of a photon at the peak wavelength emitted by the Sun (500 nm) is found using the equation
\( v = \frac{c}{\lambda} \)
where \(v\) is the frequency, \(c\) is the speed of light (approximately \(3.00 \times 10^8 \frac{m}{s}\)), and \(\lambda\) is the wavelength. When the wavelength of 500 nm (\(5.00 \times 10^{-7} m\)) is plugged into the equation, we determine the photon frequency to be \(6.00 \times 10^{14} Hz\), a value occurring in the visible spectrum which humans perceive as light.
Photon Energy
Photon energy is directly related to the frequency; the higher the frequency, the greater the energy. The equation that expresses this relationship is
\( E = h \cdot v \)
where \(E\) represents the photon energy, \(h\) is Planck's constant (\(6.63 \times 10^{-34} J\cdot s\)), and \(v\) is the frequency. Using our calculated frequency of \(6.00 \times 10^{14} Hz\), we find the energy of each photon to be \(3.98 \times 10^{-19} J\). This energy is indicative of a photon in the visible spectrum and is fundamental to understanding how sunlight interacts with objects, such as solar panels or the human retina.
Number Flux of Photons
The number flux of photons describes how many photons pass through a certain area every second. It's a convenient way to measure the intensity of sunlight reaching Earth. To find it, we divide the Solar Constant, which represents the radiant energy received from the Sun per unit area, by the energy per photon.
\( Flux = \frac{\text{Solar Constant}}{E} \)
With the Solar Constant at \(1400 \frac{W}{m^2}\) and the photon energy previously found, we calculate the number flux to be \(3.52 \times 10^{21} \frac{photons}{m^2\cdot s}\). This flux helps us estimate the rate at which photons are reaching and potentially providing energy to the Earth's surface.
Planck's Constant
Planck's constant (\(h\)) is a fundamental piece in quantum mechanics, inferring that energy transfer occurs in discrete amounts called quanta. Its current accepted value is \(6.63 \times 10^{-34} J\cdot s\), and it shows up in several essential quantum equations, including those for photon energy and the photoelectric effect. Planck’s constant reflects the granular nature of energy at microscopic scales, which debunks the classical notion of energy as a continuous variable. The tiny number represents the scale at which quantum effects become noticeable. It also illustrates that light, although often thought of as a wave, can also behave as a particle (photon), with discrete energy related to its frequency.

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

Why is a white-hot object hotter than a red-hot object?

An accelerator boosts a proton's kinetic energy so that the de Broglie wavelength of the proton is \(3.5 \cdot 10^{-15} \mathrm{~m}\) What are the momentum and energy of the proton?

Scintillation detectors for gamma rays transfer the energy of a gamma-ray photon to an electron within a crystal, via the photoelectric effect or Compton scattering. The electron transfers its energy to atoms in the crystal, which re-emit it as a light flash detected by a photomultiplier tube. The charge pulse produced by the photomultiplier tube is proportional to the energy originally deposited in the crystal; this can be measured so an energy spectrum can be displayed. Gamma rays absorbed by the photoelectric effect are recorded as a photopeak in the spectrum, at the full energy of the gammas. The Compton-scattered electrons are also recorded, at a range of lower energies known as the Compton plateau. The highest-energy of these form the Compton edge of the plateau. Gamma-ray photons scattered \(180 .^{\circ}\) by the Compton effect appear as a backscatter peak in the spectrum. For gamma-ray photons of energy \(511 \mathrm{KeV}\) calculate the energies of the Compton edge and the backscatter peak in the spectrum.

X-rays having energy of 400.0 keV undergo Compton scattering from a target. The scattered rays are detected at \(25.0^{\circ}\) relative to the incident rays. Find a) the kinetic energy of the scattered \(\mathrm{X}\) -ray, and b) the kinetic energy of the recoiling electron.

Consider an object at room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right)\) and the radiation it emits. For radiation at the peak of the spectral energy density, calculate a) the wavelength, c) the energy of one b) the frequency, and photon.

See all solutions

Recommended explanations on Physics Textbooks

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.