Energy of a Photon

\[E = h\!f = \frac{hc}{\lambda}\quad[\text{Eq. }(8.1)]\quad\text{or}\quad E_\text{eV} = \frac{1240}{\lambda_\text{nm}}\text{ eV}\quad[\text{Eq. }(8.3)]\]

Note: If you feel uncomfortable with working with variables, fractions, exponents, prefixes, or scientific notation, you should consider reviewing one or more of the following tutorials before working through this tutorial:

Sections

Introduction

Planck introduced his equation, \(E = h\!f = hc/\lambda\) in 1900 to describe discrete amounts of light energy required to mathematically derive an equation that produces the observed graph of intensity as a function of wavelength (or frequency) for blackbody radiation. The constant he created, \(h\), to match the blackbody data is known today as Planck’s constant, and is one of the fundamental constants of nature. It was Einstein who realized that Planck’s energy equation was actually specifying the energy of individual light particles that have a frequency \(f\) or a wavelength \(\lambda\), where the frequency and wavelength are related through the speed of light by \(c = f\lambda\). Einstein also proposed an experiment to test his hypothesis, referred to as the photoelectric effect, which verified the particle nature of light. Today, those particles of light are known as photons.

Planck’s solution for the blackbody radiation curve and Einstein’s explanation of the key equation [(Equation (8.1)], along with the introduction of Planck’s constant, represented the beginning of a scientific revolution in our understanding of light and of matter on the smallest scales. The overall theory is called quantum mechanics, and it is joined by general relativity as the two pillars on which all of physics is ultimately based.

The first form of Planck’s equation on the top of the page [(Equation (8.1)] requires all quantities be expressed in standard SI units (kg, m, s), while the second version [Equation (8.3)] requires that the wavelength be expressed in nanometers (nm), giving the energy of the photon in eV. Equation (8.3) results from expressing the combined constants \(hc\) in units of nanometers for length and electron volts (eV) for energy, where \(1\text{ eV} = 1.6\times 10^{-19}\text{ J}\) (one joule is \(1\text{ J} = 1\text{ N}\,\text{m} = 1\text{ kg}\,\text{m}^2/\text{s}^2\) and is the SI unit for energy). One electron volt is the kinetic energy that an electron would have after being accelerated through one volt (the same volt unit used for batteries and electric circuits). An electron volt is a convenient quantity of energy because it is typical of the energies associated with electrons.

Examples

  1. What is the energy of a photon that has a wavelength of 400 nm? What wavelength region is this?

    Using Equation (8.3), \(E_\text{eV} = \frac{1240}{400}\text{ eV} = 3.1\text{ eV}.\) This corresponds to the shortest-wavelength blue/violet light or the longest-wavelength ultraviolet light.
  2. Based on the answer for Example 1, what is the energy of a photon having a wavelength of 100 nm? What wavelength region does this fall in?

    Since 100 nm is one-fourth of the length of 400 nm, substituting into Equation (8.3) would mean that you would be dividing by a number that is smaller by a factor of 4, and so the result must be greater by a factor of 4, so \(E_\text{eV} = 4 \times 3.1\text{ eV} = 12.4\text{ eV}.\) This is an ultraviolet photon.
  3. What is the energy of a photon having a wavelength of 800 nm? And the wavelength region?

    Realizing that 800 nm is twice as long as 400 nm, the energy of the 800 nm photon must be one-half the energy of the 400 nm photon, or \(3.1\text{ eV}/2 = 1.55\text{ eV}.\) An 800 nm photon is an infrared photon.
  4. What is the energy of a \(10\ \mu\text{m}\) photon?

    From Table 5.1 (p. 155) or the prefixes tutorial, the prefix \(\mu\) is shorthand for \(10^{-6}\) and n is shorthand for \(10^{-9}\). As a result, \(10\ \mu\text{m} = 10\times10^{-6}\text{ m} = 10\times10^3\times10^{-9}\text{ m} = \text{10,000 nm}\). This means that \(\lambda_\text{nm} = \text{10,000}\), and so \[E_\text{eV} = \frac{1240}{\text{10,000}}\text{ eV} = 0.124\text{ eV}.\]
  5. A photon has an energy of \(5\text{ MeV}\). What is the photon’s wavelength?

    Remember that if you multiple both sides of an equation by the same number, the equality remains true: for example, if \(2 = 2\) then \(3 \times 2 = 3 \times 2\), otherwise written as \(6 = 6\). The same is true of variables; for example, if \(x = y\), then \(3x = 3y\), or if \(x = y\), then \(ax = ay\). Don’t forget that algebra may appear abstract, but in reality it is just generalized arithmetic.

    In order to determine the photon’s wavelength, Equation (8.3) must be solved from \(\lambda_\text{nm}\). This can be accomplished by multiplying both sides by the same quantities, in this case \(\frac{\lambda_\text{nm}}{E_\text{eV}}.\) By doing so, we have \[\frac{\lambda_\text{nm}}{E_\text{eV}} \times E_\text{eV} = \frac{\lambda_\text{nm}}{E_\text{eV}} \times \frac{1240}{\lambda_\text{nm}}.\] Since \(E_\text{eV}\) is in both the numerator and the denominator on the left-hand side, it cancels. Similarly, since \(\lambda_\text{nm}\) is in the numerator and the denominator on the right-hand side, it cancels as well. The result is \[\lambda_\text{nm} = \frac{1240}{E_\text{eV}}\text{ nm}.\]

    From Table 5.1 or the prefixes tutorial, \(5\text{ MeV} = 5\times10^6\text{ eV},\) so \(E_\text{eV} = 5\times10^6.\) Substituting, \[\lambda_\text{nm} = \frac{1240}{5\times10^6}\text{ nm} = 0.000\,248\text{ nm} = 0.248\text{ pm} = 248\text{ fm}.\] This result is also \(2.48\times 10^{-13}\text{ m}.\)
  6. A photon has an energy of 13.6 eV. What is its wavelength in nanometers? What wavelength region is it in?

    Using the equation from Example 5: \[\lambda_\text{nm} = \frac{1240}{13.6}\text{ nm} = 91.2\text{ nm}.\] The wavelength falls in the ultraviolet portion of the electromagnetic spectrum.

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

  1. A photon has a wavelength of 550 nm. What is the photon’s energy, and what part of the electromagnetic spectrum does it correspond to?
  2. A photon has a wavelength of 2 nm. What is the photon’s energy, and what part of the electromagnetic spectrum does it corresponds to?
  3. An important wavelength for studying the structure of our Galaxy is 21.1 cm. Express that wavelength in nm. What is the energy of a 21.1 cm photon? What is the photon’s frequency?
  4. A radio station broadcasts with a wavelength of 492 m. What is the energy of the emitted photons?
  5. Carbon atoms can emit photons with a wavelength of 601 nm. What are the photons’ energies?
  6. The atoms of a particular element are found to produce photons with an energy of 2.55 eV. What wavelength does that energy correspond to? Referring to Table 7.2 (page 252) in your textbook, what element is the sample likely to be?

(Answers are available below.)

Answers
  1. 2.25 eV. The wavelength is in the middle of the visible spectrum.
  2. 620 eV. It is an x-ray photon.
  3. \(\lambda_\text{nm} = 2.1\times 10^8\text{ nm}\). \(E_\text{eV} = 5.92\times10^{-5}\text{ eV} = 59.2\,\mu\text{eV}\). \(f = 1420\text{ MHz}\)
  4. \(2.52\text{ neV} = 2.52 \times 10^{-9}\text{ eV}\)
  5. \(2.06\text{ eV}\)
  6. \(486\text{ nm}\). Hydrogen
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