Answer :
Final answer:
Using Wien's Law with the peak wavelength of 483 nm, the temperature of the star is calculated to be approximately 6000.62 K. The closest given option is C 6,180 K, which is the answer to the question.
Explanation:
To find the temperature of a star that emits most of its light at 483 nm, we can use Wien's Law, which relates the wavelength of the peak emission of a blackbody to its temperature. The law is expressed as λmax T = b, where λmax is the peak wavelength in meters, T is the temperature in kelvins, and b is Wien's displacement constant (approximately 2.898 x 10-3 m·K).
First, we convert the given wavelength from nanometers to meters: 483 nm = 483 x 10-9 m. Then, we rearrange Wien's Law to solve for T: T = b / λmax. Plugging in the values gives us T = (2.898 x 10-3 m·K) / (483 x 10-9 m) = 6000.62 K. However, none of the given options exactly match this result, so we choose the one that is closest. In this case, option c) 6,180 K is the closest to our calculated temperature. Therefore, the correct answer is 6,180 K.
