Wien’s Displacement Law (DP IB Physics: SL): Revision Note

The black-body radiation curve of an object at 900 K is shown in the diagram below.

Which of the following shows the black-body radiation curve of an object at 650 K?

The dashed line represents the curve of the object at 900 K.

Answer:  C

• From Wien's displacement law:  ${\lambda }_{max}\propto \frac{1}{T}$

• Therefore, a curve with a longer peak wavelength will correspond to a lower temperature

The spectrum of the star Rigel in the constellation of Orion peaks at a wavelength of 263 nm, while the spectrum of the star Betelgeuse peaks at a wavelength of 828 nm.

Determine which of these two stars, Betelgeuse or Rigel, is cooler.

Answer:

Step 1: List the known quantities

• Maximum emission wavelength of Rigel = 263 nm = 263 × 10⁻⁹ m

• Maximum emission wavelength of Betelgeuse = 828 × 10⁻⁹ m

Step 2: Write down Wien’s displacement law and rearrange for temperature T

${\lambda }_{max}T=2.9\times {10}^{-3}\mathrm{m}\mathrm{K}$

$T=\frac{2.9\times {10}^{-3}}{{\lambda }_{max}}$

Step 3: Calculate the surface temperature of each star

Rigel:  $T=\frac{2.9\times {10}^{-3}}{263\times {10}^{-9}}=11027=11000\mathrm{K}$

Betelgeuse:  $T=\frac{2.9\times {10}^{-3}}{828\times {10}^{-9}}=3502=3500\mathrm{K}$

Step 4: Write a concluding sentence

• Betelgeuse has a surface temperature of 3500 K, therefore, it is much cooler than Rigel

Wien's law and stars in the Orion constellation

The Orion Constellation; cooler stars, such as Betelgeuse, appear red or yellow, while hotter stars, such as Rigel, appear white or blue

Note that the temperature used in Wien’s Law is in Kelvin (K). Remember to convert from ^oC if the temperature is given in degrees in the question before using the Wien’s Law equation.