A given star has a peak emission wavelength of 60nm, lies 7.10*10^19m away and the intensity of its electromagnetic radiation reaching the Earth is 3.33*10^-8Wm^-2. Calculate the star's diameter
With a problem like this, the key is to split it down into component parts.
We will treat the star as a perfect emitter and radiator, something known as a black body. There will be two physical laws we need to use:
-Stefan-Boltzmann law: P=σAT^4 where P=power dissipated by a black body, σ=Stefan-Boltzmann constant, 5.6710^-8 W(m^-2)(K^-4), A=surface area of the body, T=temperature
-Wien's law: λmax=W/T where λmax=peak emission wavelength, W=Wien's constant, 2.9010^-3 Km, T=temperature
Step 1: Finding the star's temperature
The peak emission wavelength of the star is given in the question as 60nm, which is 6.010^-8 m in standard form. Re-arranging the formula for Wien's law we get:
T=λmax/W
T=(6.010^-8)/(2.9010^-3)
T=48330 K 4.s.f
Step 2: Finding the power of the star
In order for us to use the Stefan-Boltzmann law, we need the power emitted by the star. Currently we have the intensity at the Earth's surface. Light propagates out spherically so the intensity is given by:
I=P/(4πr^2) where r=distance from star to Earth
Re-arranging this, we get:
P=4πIr^2
P=4π(3.3310^-8)(7.1010^19)^2
P=2.10910^33 W 4.s.f
Step 3: Finding the surface area of the star
Re-arranging the Stefan-Boltzmann law we get:
A=P/(σT^4)
A=(2.10910^33)/(5.6710^-8)(48330)^4
A=6.81810^21 m^2 4.s.f
Step 4: Finding the diameter of the star
As the star is spherical, it's area is 4πr^2, that is πd^2. Re-arranging this we get:
d=sqrt(A/π)
d=sqrt(6.81810^21/π)
Diameter= 4.66*10^10 m 3.s.f
Note on significant figures: By making sure to keep to 4.s.f at each stage of the calculation, you ensure that the final answer will be correct to 3.s.f
