The light from the sun is found to have a maximum intensity near the wavelength of 470nm. Assuming that the surface o f the sun emits as a blackbody, calculate the temperature of the surface of the sun.

Explanation

Let the diameter of the sun be D and its distance from the earth be R. From the question,
`D/R~~0.53xxpi/180` . `9.25xx10^(-3)` . The radiation emitted by the surface of the sun per unit time is
`4pi((D)/(2))^(2)sigmaT^(4)=piD^(2)sigmaT^(4)` . At distance R, this radiation falls on an area `4piR^(2)` . in unit time. The radiation received at the earth's surface per unit time per unit area is, therefore,
`(piD^(2)sigmaT^(4))/(4piR^(2))=(sigmaT^(4))/(4)(D/R^(2))` . thus, `(sigmaT^(4))/(4)(D/R)^(2)=8.2Jcm^(-2)min^(-1)` . or, `(1)/(4)xx(5.67xx10^(-s)Wm^(-2)K^(-4))T^(4)xx(9.25xx10^(-2))^(2)` . `=(8.2)/(10^(-4)xx60)Wm^(-2)` . or, `T=5794K~~5800K` .