Question

Wien's law states that the peak wavelength is inversely proportional to the temperature. if star A is hotter (i.e. higher temperature) than star B, then star A will peak at a wavelength that is _____________ the peak wavelength of the spectrum of star B

Answer 1

If star A is hotter (i.e. higher temperature) than star B, then star A will peak at a wavelength that is SHORTER THAN the peak wavelength of the spectrum of star B

Step-by-step explanation:

Wien's law states that the peak wavelength is inversely proportional to the temperature. That is, at increased temperature the wavelength of peak emission will be at a shorter wavelength. Frome the example of stars given, star A has a higher temperature than star B, therefore it's wavelength will peak at a wavelength that is shorter than the peak wavelength of the spectrum of star B.

Answer 2

The answer is: star A with higher temperature would peak at a wavelength that is lower in the electromagnetic spectrum than the peak wavelength of star B.

Step-by-step explanation:

Wien's law gives us the correlation between the temperature of a black body (an ideal body which perfectly absorbs or emits radiation at all frequencies) and the wavelength at which it emits the light.

A plot of electromagnetic energy vs. temperatures indicates that the peak intensity and wavelength of emission is inversely proportional to the temperature,hence the maximum shifts to shorter wavelengths occurs as the temperature T is increased.

Taking these into consideration, star A with higher temperature would peak at a wavelength that is lower in the electromagnetic spectrum than the peak wavelength of star B.