Star A has the peak of its blackbody radiation at $\lambda$_A. Star B has its peak at $\lambda$_B, which is one-third that of $\lambda$_A. If Star A's surface temperature is T_A, how does the surface temperature T_B of Star B compare? A) T_B = T_A/9 B) T_B = 3 T_A C) T_B = 9 T_A D) T_B = T_A/3

Question

Quizplus · Accepted Answer

According to Wien's Law, the peak wavelength is inversely proportional to the temperature. If Star B's peak wavelength is one-third of Star A's, then Star B's temperature is three times that of Star A.

Star a Has the Peak of Its Blackbody Radiation At \lambda