TP 30.8

Given: T = 2900 K

Find: (a) f _peak- ?

f _peak = (5.88 x 10¹⁰ s^-1K^-1) T

f _peak = (5.88 x 10¹⁰ s^-1K^-1) (2900 K) = 1.71 x 10¹⁴ Hz

(b) Since the peak frequency is well into the infrared region of the spectrum, the light bulb radiates more energy in the infrared part of the spectrum.

(Note that we seldom use ordinary light bulbs for heating purposes, so the answer indicates that we waste a lot of energy generating heat we cannot put to a usefull purpose.)