Answer: The star Alnilam is hotter
Explanation:
The question seems to be the continuation of another question, which is written below:
Use Wien's law to calculate the surface temperature of the star Alnilam in Kelvin. The peak wavelength [tex]\lambda_{p}[/tex] of the star is [tex]0.116 \mu m[/tex]
So, by using Wien's displacement law, which relates the wavelength [tex]\lambda_{p}[/tex] where the intensity of the radiation is maximum with the temperature [tex]T[/tex] of the black body, and expresed as:
[tex]\lambda_{p}.T=C[/tex] (1)
Where:
[tex]T[/tex] is in Kelvin (K)
[tex]\lambda_{p}=0.116 \mu m=0.116 (10)^{-6} m[/tex] is the wavelength of the emission peak in meters (m).
[tex]C=0.0028976 mK[/tex] is the Wien constant
We can find Alnilam's temperature and then compare it with the Sun's temperature ([tex]T_{Sun}=5778 K[/tex]):
[tex]T_{Alnilam}=\frac{C}{\lambda_{p}}[/tex] (2)
[tex]T_{Alnilam}=\frac{0.0028976 mK}{0.116 (10)^{-6} m}[/tex] (3)
[tex]T_{Alnilam}=24979.310 K[/tex] (4)
As we can see [tex]T_{Alnilam} > T_{Sun}[/tex], hence:
