Question

In the figure, the distribution of energy density of the radiation emitted by a black body at a given temperature is shown. The possible temperature of the black body is

A. 1500 K
B. 2000 K
C. 2500 K
D. 3000 K

Answer 1

Correct Answer - B
`lambda_(m)T`=b, where b = `2.890 xx 10^(-3)`mK
`rArr T=b/lambda_(m)` = (2.89 xx 10^(-3))/(1.5 xx 10^(-6))` = 2000K