Question

An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody emitters. For each star she makes a plot of the radiation intensity per unit wavelength as a function of wavelength. She notices that the curve for star A has a maximum that occurs at a shorter wavelength than does the curve for star B. What can she conclude about the surface temperatures of the two stars

Answer 1

Star A has a higher surface temperature than star B.

Step-by-step explanation:

The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:

`T = (2.898x10^(-3) m. K)/(\lambda max)` (1)

Where T is the effective temperature of the star and
`\lambda_(max)` is the maximum peak of emission.

A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.

Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and
`\lambda max` are inversely proportional.