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Chapter 38: Problem 3

A photon of green light has a wavelength of 520 nm. Find the photon's frequency, magnitude of momentum, and energy.Express the energy in both joules and electron volts.

Short Answer

Expert verified
Frequency: \(5.77 \times 10^{14}\, \text{Hz}\), Energy: \(3.82 \times 10^{-19}\, \text{J}\) or \(2.38\, \text{eV}\), Momentum: \(1.27 \times 10^{-27}\, \text{kg} \cdot \text{m/s}\).

Step by step solution

01

Convert Wavelength to Meters

The wavelength of the photon is given as 520 nm. To perform calculations, convert this to meters by using the conversion: \[ \text{1 nm} = 10^{-9} \, \text{m} \]Thus, the wavelength \( \lambda \) in meters is:\[ \lambda = 520 \, \text{nm} = 520 \times 10^{-9} \, \text{m} = 5.2 \times 10^{-7} \, \text{m} \]
02

Calculate the Frequency

Use the speed of light formula to find the frequency (\( u \)): \[ c = \lambda u \]where \( c = 3 \times 10^8 \, \text{m/s} \). Solving for frequency gives:\[ u = \frac{c}{\lambda} = \frac{3 \times 10^8}{5.2 \times 10^{-7}} = 5.77 \times 10^{14} \, \text{Hz} \]
03

Calculate the Energy in Joules

The energy of a photon is given by Planck's equation: \[ E = h u \]where \( h = 6.626 \times 10^{-34} \, \text{Js} \) is Planck's constant. Substituting the frequency:\[ E = 6.626 \times 10^{-34} \times 5.77 \times 10^{14} = 3.82 \times 10^{-19} \, \text{J} \]
04

Convert Energy to Electron Volts

To convert energy from joules to electron volts, use the conversion:\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]Thus, the energy in electron volts is:\[ E = \frac{3.82 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.38 \, \text{eV} \]
05

Calculate the Momentum

The momentum (\( p \)) of the photon can be calculated using:\[ p = \frac{E}{c} \]Substituting the energy in joules and the speed of light:\[ p = \frac{3.82 \times 10^{-19}}{3 \times 10^8} = 1.27 \times 10^{-27} \, \text{kg} \cdot \text{m/s} \]

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Key Concepts

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

Wavelength Conversion
Wavelength conversion is an essential step in photon energy calculations, especially when dealing with nanometers (nm), a common unit for measuring wavelengths of light. To convert nanometers to meters, which is the standard unit in physics, use the relation:
  • 1 nanometer (nm) = 10-9 meters (m)
This conversion is crucial because physical equations like those involving speed of light or energy require wavelengths to be in meters. For instance, a wavelength of 520 nm becomes 520 脳 10-9 meters, resulting in 5.2 脳 10-7 meters. This conversion helps maintain consistency across calculations.
Frequency Calculation
Finding the frequency of a photon involves using the basic relationship between wavelength and frequency dictated by the speed of light. The speed of light (\[ c \]) is a constant, approximately 3 脳 108 meters per second. The formula connecting wavelength (\[ \lambda \]), frequency (\[ u \]), and speed of light is:
  • \[ c = \lambda u \]
To solve for frequency, rearrange the equation to:
  • \[ u = \frac{c}{\lambda} \]
Substitute the given values to calculate frequency. For our 520 nm wavelength (5.2 脳 10-7 m), the calculation would be:
  • \[ u = \frac{3 \times 10^8}{5.2 \times 10^{-7}} = 5.77 \times 10^{14} \text{ Hz} \]
This frequency tells us how many wave peaks pass a point in one second. Understanding this concept helps in various fields, including communications and optics.
Momentum of Photon
Photon momentum seems counterintuitive because photons are massless particles. However, they do carry momentum, evident through interactions like hitting surfaces and transmitting energy. The momentum (\[ p \]) of a photon is derived from its energy divided by the speed of light:
  • \[ p = \frac{E}{c} \]
Using Planck's constant, we calculate photon energy (\[ E \]) with the equation \[ E = hu \] (where \[ h \] is Planck's constant). Given that the energy has already been calculated as 3.82 脳 10-19 J, divide by the speed of light:
  • \[ p = \frac{3.82 \times 10^{-19}}{3 \times 10^8} = 1.27 \times 10^{-27} \text{ kg} \cdot \text{m/s} \]
This value indicates the momentum associated with the photon, important in understanding light's interaction with matter, such as exerting pressure on a solar sail.
Energy Conversion to Electron Volts
Energy conversion to electron volts (eV) is particularly useful in physics and engineering, where energies involved are often very small when expressed in joules. One electron volt is the amount of energy a single electron gains moving across an electric potential difference of one volt. The conversion factor is:
  • 1 eV = 1.602 脳 10-19 joules (J)
To convert our calculated photon energy from joules (3.82 脳 10-19 J) into electron volts, use:
  • \[ E = \frac{3.82 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.38 \text{ eV} \]
This conversion is practical, making it easier to handle numbers when discussing energy levels in atomic physics or semiconductor devices.

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Most popular questions from this chapter

X Rays from Television Screens. Accelerating voltages in cathode-ray-tube (CRT) TVs are about 25.0 \(\mathrm{kV}\) . What are (a) the highest frequency and (b) the shortest wavelength (in nm) of the x rays that such a TV screen could produce? (c) What assumptions did you need to make? (CRT televisions contain shielding to absorb these x rays.)

An incident x-ray photon of wavelength 0.0900 \(\mathrm{nm}\) is scattered in the backward direction from a free electron that is initially at rest. (a) What is the magnitude of the momentum of the scattered photon? (b) What is the kinetic energy of the electron after the photon is scattered?

(a) A proton is moving at a speed much slower than the speed of light. It has kinetic energy \(K_{1}\) and momentum \(p_{1}\) . If the momentum of the proton is doubled, so \(p_{2}=2 p_{1},\) how is its new kinetic energy \(K_{2}\) related to \(K_{1} ?(b)\) A photon with energy \(E_{1}\) has momentum \(p_{1} .\) If another photon has momentum \(p_{2}\) that is twice \(p_{1},\) how is the energy \(E_{2}\) of the second photon related to \(E_{1} ?\)

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The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the \(n=2\) level to the \(n=1\) level?
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