Note: If you feel uncomfortable with working with scientific notation, variables, or fractions, you should review the following tutorials before working through this tutorial:
Introduction
Wien’s blackbody color law [Equation (6.12)] and the equation for determining the temperature from the blackbody peak wavelength [Equation (6.13)] are directly related to one another (the second law is just Wien’s law solved for surface temperature) and so will be treated together in this tutorial.
Sections
Wien’s Blackbody Color Law
$$\lambda_\text{peak} = \frac{\text{2,900,000 nm K}}{T_\text{surface}}\quad[\text{Eq. (6.12)}]$$
Wien’s law is a fairly straightforward equation to apply. The number on top of the fraction (the numerator) is simply a constant that never changes. Ultimately the constant is based on a group of fundamental constants, like the speed of light, \(c\). This means that the only two variables in the equation are the peak wavelength of the blackbody radiation curve [see Fig. (6.43)] and the surface temperature of the blackbody, such as a star or a planet. The constant does make explicit the units that need to be used in the equation, however. The length unit that must used for the blackbody curve’s peak wavelength, \(\lambda_\text{peak}\), is nanometers (nm, where \(1~\text{nm} = 1\times 10^{-9}~\text{m}\). The surface temperature must be expressed in kelvins, \(\text{K}\). If the temperature you are given is in Celsius (\({}^\text{o}\text{C}\)) or Fahrenheit (\({}^\text{o}\text{F}\)) for example, then the temperature must be converted to kelvins before putting it into Wien’s law; see Equation (6.11).
The behavior of Wien’s law is identical the equation, \(y = 1/x\). If you substitute \(1\) in for \(x\), then the result is \(y = 1/1 = 1\). If you double the value of \(x\), the value for \(y\) is now one-half as large, or \(y = 1/2\). Increasing \(x\) to \(5\) makes \(y\) become one-fifth as large as it was when \(x\) was \(1\). Reducing \(x\) to values less than \(1\), but still greater than zero, has the effect of increasing the value of \(y\). For example, if the value of \(x\) is one-half its original value, then $$y = \frac{1}{\frac{1}{2}} = \frac{1}{\frac{1}{2}} \times 1 = \frac{1}{\frac{1}{2}} \times \frac{2}{2} = \frac{1\times 2}{\frac{1}{2}\times 2} = \frac{2}{\frac{2}{2}} = \frac{2}{1} = 2.$$ You need not go through all of those steps as long as you realize that \(\frac{1}{1/z} = z\).
Examples:
1. The Sun’s surface temperature is \(5772~\text{K}\). What is its peak blackbody wavelength?
Substituting into Wien’s law gives $$\lambda_\text{peak} = \frac{\text{2,900,000 nm K}}{5772~\text{K}} = \frac{\text{2,900,000}}{5772} \text{nm} = 502~\text{nm}.$$ Note that since the temperature unit (\(\text{K}\)) was in the numerator and the denominator, it could simply be canceled.
2. Suppose that a star has a surface temperature that is exactly twice as hot as the surface temperature of the Sun, \(\text{11,544 K}\). What is its peak blackbody wavelength?
Doing what we did in the first example: $$\lambda_\text{peak} = \frac{\text{2,900,000 nm K}}{\text{11,544 K}} = 251~\text{nm}.$$ This result could also have been obtained by realizing that \(\text{11,544 nm} = 2\times\text{5772 nm}\), and so $$\lambda_\text{peak} = \frac{\text{2,900,000 nm K}}{\text{11,544 K}} = \frac{\text{2,900,000 nm K}}{\text{5772 K}\times 2} = \frac{\text{2,900,000 nm K}}{5772~\text{K}} \times \frac{1}{2} = (\text{502 nm})\times \frac{1}{2} = \text{251 nm}.$$ All that the math above is saying is that since the star’s surface temperature is twice a great as the Sun, its peak wavelength is \(1/2\) as long. [Remember \(y = 1/x\) above; if you double \(x\) you reduce \(y\) by \(1/2\). Conversely, if the star’s surface temperature is a factor of \(2\) cooler than the Sun’s, then its peak wavelength is twice as long.]
If you remember the “trick” used before in this textbook (think of Kepler’s third law), the constant \(\text{2,900,000 nm K}\) can be canceled if the general Wien’s law equation is divided by the equation with the Sun’s values in it. This gives $$\frac{\lambda_\text{peak}}{\text{502 nm}} = \frac{1/T_\text{surface}}{1/\text{5772 K}} = \frac{\text{5772 K}}{T_\text{surface}}.$$ This last expression can also be written in another way by dividing the numerator and the denominator by \(\text{5772 K}\) which gives: $$\lambda_\text{peak} = \frac{\text{502 nm}}{T_\text{surface}/\text{5772 K}}\quad[\text{this is just another form of Wien’s law}.]$$
More Examples:
3. Suppose that a star has a surface temperature of \(\text{3000 K}\). What is its peak blackbody wavelength?
Since \(\text{3000 K}/\text{5772 K} = 0.52\), the peak wavelength is \(\lambda_\text{peak} = \text{502 nm}/0.52 = \text{965 nm}\).
4. The average temperature of a planet orbiting another star is \(\text{1200 K}\). What is its peak blackbody wavelength?
\(\text{1200 K}/\text{5772 K} = 0.208\), so \(\lambda_\text{peak} = \text{502 nm}/0.208 = \text{2413 nm}.\)
5. You radiate much like a blackbody, but at a temperature of about \(\text{97}^\text{o}\text{F}\), which is \(\text{36}^\text{o}\text{C}\) or \(\text{309 K}\). What is your peak wavelength?
\(\text{309 K}/\text{5772 K} = 0.054\). As a result, your peak blackbody wavelength is \(\lambda_\text{peak} = \text{502 nm}/0.054 = \text{9300 nm}\).
Temperature From the Peak Blackbody Wavelength
Wien’s blackbody color law can be rewritten by solving for the surface temperature if the peak wavelength is known. In this form
$$T_\text{surface} = \frac{\text{2,900,000 nm K}}{\lambda_\text{peak}}\quad[\text{Eq. (6.13)}].$$
Using the alternate form of Wien’s law, the surface temperature can also be expressed as
$$T_\text{surface} = \frac{\text{5772 K}}{\lambda_\text{peak}/\text{502 nm}}.$$
Examples:
6. A star has a peak blackbody wavelength that is one-third of the Sun’s peak wavelength. What is its surface temperature?
\(T_\text{surface} = \text{5772 K}/(1/3) = 3\times \text{5772 K} = \text{17,316 K}\).
7. The peak wavelength of a very cool star is found to be \(\text{1200 nm}\). What is its surface temperature?
$$T_\text{surface} = \text{5772 K}/(\text{1200 nm}/\text{502 nm}) = \text{5772 K}/\text{2.39} = \text{2414 K}$$.
Using the original form in Equation (6.13),
$$T_\text{surface} = \frac{\text{2,900,000 nm K}}{\lambda_\text{peak}} = \frac{\text{2,900,000 nm K}}{\text{1200 nm}} = \text{2417 K}.$$ The difference between the two results is simply not being highly precise in the numbers used (i.e., round-off errors).
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.
(Answers are available below.)
- A very distance dwarf planet has a surface temperature of \(\text{12 K}\). (a) What fraction of the Sun’s surface temperature is it? (b) What is its peak blackbody wavelength? (c) According to Table (6.1), what portion of the electromagnetic spectrum does the wavelength correspond to?
- Star A has a surface temperature of \(\text{3000 K}\) and star B has a surface temperature of \(\text{12,000 K}\). (a) What is star A’s peak blackbody wavelength? What wavelength region does that correspond to? (b) Star B’s peak wavelength is how many times greater or less than star A’s wavelength? What wavelength region does that result correspond to?
- During a star’s lifetime its surface temperature may go from \(\text{10 K}\) to \(\text{40,000 K}\). (a) What are the peak wavelengths at those to points in the star’s life? (b) Which portions of the electromagnetic spectrum do they belong to?
- Determine the peak wavelength of a planet that has a surface temperature of \(\text{700 K}\).
- Studying the blackbody spectrum of a distant star reveals that it has a peak wavelength of \(\text{200 nm}\). What surface temperature does that correspond to?
- The enormous star, Betelgeuse, in the shoulder of the constellation of Orion, has a peak wavelength in the infrared portion of the electromagnetic spectrum. Would you expect its surface temperature to be closer to \(\text{3000 K}\), \(\text{5000 K}\), \(\text{10,000 K}\), or \(\text{20,000 K}\)?
Answers
- (a) \(\text{0.0021}\), or \(\text{0.21}\%\). (b) \(\text{241,000 nm} = \text{0.241 mm}\). (c) Very far into the infrared.
- (a) \(\text{965 nm}\). Infrared (b) B has four times the surface temperature of A and therefore a peak wavelength that is one-fourth as long, or \(\text{241 nm}\). Ultraviolet.
- (a) At \(\text{10 K}\), the peak wavelength is \(\text{290,000 nm} = 2.9\times 10^{-4}~\text{m} = \text{0.29 mm}\). At \(\text{40,000 K}\) the peak wavelength is \(\text{72 nm}\). (b) At \(\text{10 K}\) the peak wavelength is in the infrared. At \(\text{40,000 K}\) the peak is in the ultraviolet.
- \(\text{4130 nm} = \text{4.13}~\mu\text{m}\)
- \(\text{14,500 K}\).
- \(\text{3000 K}\).