The Sun's surface has a temperature of 5800K. What will...?

Question

... what will be the peak wavelength of the spectrum of a star that emits twice the sun's flux of energy?

a) 2.1 x 10e-7 m
b) 4.0 x 10e-7 m
c) 5.0 x 10e-7 m
d) 1.0 x 10e-6 m

I know I have to use Fsun = r T^4 then use Wein's law. but my answer doesn't match any of the above answer. i somehow got 1.45 x 10e-3

Answer 1

You don't need to use r T^4, you simply use Wien's law. You are given that T=5800. Wien's law says that (lambda)*T=2.898*10^(-3), so just divide 2.898*10^(-3) by 5800. I get c) as the answer.

Answer 2

A star emitting twice the Sun's power will have a temperature of 2^(1/4) times that of the Sun. That is 5800 times1.189=6897K. The peak wavelength times the temperature is constant. So the new peak wavelength will be that of the Sun times 5800/6897.

Answer 3

The solar will emit its height radiation at a shorter wavelength than the cooler celebrity. you ought to use Wien's (displacement) regulation to unravel for exact values, yet, warmer products emit mild at shorter (bluer) wavelengths.

Answer 4

see:
http://en.wikipedia.org/wiki/Wien%27s_Displacement_Law