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A photovoltaic device uses monochromatic light of wavelength 700 . \(\mathrm{nm}\) that is incident normally on a surface of area \(10.0 \mathrm{~cm}^{2}\). Calculate the photon flux rate if the light intensity is \(0.300 \mathrm{~W} / \mathrm{cm}^{2}\).

Short Answer

Expert verified
Answer: The photon flux rate is approximately \(1.06 \times 10^{19}\) photons per second.

Step by step solution

01

Calculate the energy of a single photon

Using the wavelength of the light, we can determine the energy of a single photon using the Plank's equation: \(E = h \cdot f\) where \(E\) is the energy of the photon, \(h\) is the Planck's constant (\(6.63 \times 10^{-34}\, \mathrm{Js}\)), and \(f\) is the frequency of the light. To find the frequency, we can use the speed of light formula: \(c = f \cdot \lambda\) where \(c\) is the speed of light (\(3 \times 10^8\, \mathrm{m/s}\)), \(f\) is the frequency, and \(\lambda\) is the wavelength. We'll rearrange this equation to solve for \(f\): \(f = \frac{c}{\lambda}\)
02

Convert the wavelength to meters

The given wavelength is in nanometers (nm), but we need to convert it to meters (m) to work with the other units in this exercise. To convert nanometers to meters, multiply by \(10^{-9}\): \(\lambda = 700\, \mathrm{nm} \cdot 10^{-9}\, \mathrm{m/nm} = 7 \times 10^{-7}\, \mathrm{m}\)
03

Calculate the frequency of the light

Now that we have the wavelength in meters, we can find the frequency of the light using the speed of light formula: \(f = \frac{3 \times 10^8\, \mathrm{m/s}}{7 \times 10^{-7}\, \mathrm{m}} \approx 4.29 \times 10^{14}\, \mathrm{Hz}\)
04

Calculate the energy of a single photon

Now, we can calculate the energy of a single photon using the calculated frequency and Planck's equation: \(E = h \cdot f = (6.63 \times 10^{-34}\, \mathrm{Js}) \cdot (4.29 \times 10^{14}\, \mathrm{Hz}) \approx 2.84 \times 10^{-19}\, \mathrm{J}\)
05

Calculate the total incident power on the surface

We are given the light intensity, \(I = 0.300\, \mathrm{W/cm}^2\). We can find the total incident power on the surface by multiplying the light intensity by the surface area: \(P = I \cdot A = (0.300\, \mathrm{W/cm}^2) \cdot (10.0\, \mathrm{cm}^2) = 3\, \mathrm{W}\)
06

Calculate the photon flux rate

Finally, we can calculate the photon flux rate by dividing the total incident power by the energy of a single photon: \(\Phi = \frac{P}{E} = \frac{3\, \mathrm{W}}{2.84 \times 10^{-19}\, \mathrm{J}} \approx 1.06 \times 10^{19}\, \mathrm{photons/s}\) The photon flux rate is approximately \(1.06 \times 10^{19}\) 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 photon energy involves understanding how energy is transferred via electromagnetic waves. Photon energy is the energy carried by a single photon, a fundamental particle of light. This energy can be calculated using Planck's equation:\[ E = h \cdot f \].Here, \( E \) represents the energy of the photon, \( h \) is Planck's constant (\(6.63 \times 10^{-34}\, \mathrm{Js}\)), and \( f \) is the frequency of the light. Knowing the frequency is crucial because it directly relates to the energy a photon carries. And since light can travel at varying frequencies and wavelengths, photon energy differs for different parts of the electromagnetic spectrum. To determine the energy of our photon, we first need to find its frequency.
Light Intensity
Light intensity refers to the power per unit area the light delivers. When light hits a surface, its intensity denotes the amount of energy it provides to that surface. Calculating this uses the formula: \[ I = \frac{P}{A} \]where \( I \) represents intensity, \( P \) is the power, and \( A \) is the area. When light intensity is measured, it's typically described in watts per square centimeter (\( \mathrm{W/cm}^2 \)).In the provided exercise, the intensity tells us how much power the light beam delivers to each square centimeter of the surface. To find the total power that hits the surface, you multiply the light intensity by the area of the surface. This comprehensive calculation offers a measurement of the total incident power that will be converted into photon output.
Frequency of Light
The frequency of light is an essential characteristic that describes how many wave peaks pass a certain point in a given time period. The relationship between the speed of light, frequency, and wavelength is given by the formula:\[ c = f \cdot \lambda \]Here, \( c \) stands for the speed of light\((3 \times 10^8\, \mathrm{m/s})\), \( f \) is the frequency, and \( \lambda \) is the wavelength. By rearranging this equation, the frequency \( f \) of the light can be found through:\[ f = \frac{c}{\lambda} \].In our calculation, the wavelength is converted from nanometers to meters to match other measurements. Subsequently, by substituting the relevant values into this equation, we can find the frequency. This step is necessary to link the wavelength of the light with its photon energy.
Planck's Equation
Planck's Equation is a fundamental equation used to calculate the energy of photons based on their frequency. It is given as:\[ E = h \cdot f \].This equation underlines the quantized nature of electromagnetic radiation, implying that energy is carried in discrete packets called quanta or photons. To employ Planck's equation correctly, it is crucial to have accurate values for both Planck's constant \( h \) and the frequency of the light \( f \). Once this equation is used, it provides the exact amount of energy each photon carries. Realizing the significance of Planck's equation helps in understanding the quantization of energy in photons and contributes to the broader field of quantum mechanics. It's particularly significant in fields like photovoltaic technology, where the energy of photons is harnessed to convert light into electricity.

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Most popular questions from this chapter

A nitrogen molecule of mass \(m=4.648 \cdot 10^{-26} \mathrm{~kg}\) has a speed of \(300.0 \mathrm{~m} /\mathrm{s}\).

Alpha particles are accelerated through a potential difference of \(20.0 \mathrm{kV}\). What is their de Broglie wavelength?

Which one of the following statements is true if the intensity of a light beam is increased while its frequency is kept the same? a) The photons gain higher speeds. b) The energy of the photons is increased. c) The number of photons per unit time is increased. d) The wavelength of the light is increased.

`A pure, defect-free semiconductor material will absorb the electromagnetic radiation incident on that material only if the energy of the individual photons in the incident beam is larger than a threshold value known as the "band-gap" of the semiconductor. The known room-temperature band-gaps for germanium, silicon, and gallium-arsenide, three widely used semiconductors, are \(0.66 \mathrm{eV}, 1.12 \mathrm{eV},\) and \(1.42 \mathrm{eV},\) respectively. a) Determine the room-temperature transparency range of these semiconductors. b) Compare these with the transparency range of \(\mathrm{ZnSe}, \mathrm{a}\) semiconductor with a band-gap of \(2.67 \mathrm{eV},\) and explain the yellow color observed experimentally for the ZnSe crystals. c) Which of these materials could be used for a light detector for the 1550 -nm optical communications wavelength?

Calculate the wavelength of a) a \(2.00 \mathrm{eV}\) photon, and b) an electron with kinetic energy \(2.00 \mathrm{eV}\).

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