# 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

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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, 5.67*10^-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, 2.90*10^-3 K*m, 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 6.0*10^-8 m in standard form. Re-arranging the formula for Wien's law we get:

T=λmax/W

T=(6.0*10^-8)/(2.90*10^-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=4π*I*r^2

P=4π(3.33*10^-8)(7.10*10^19)^2

P=2.109*10^33 W 4.s.f

Step 3: Finding the surface area of the star

Re-arranging the Stefan-Boltzmann law we get:

A=P/(σ*T^4)

A=(2.109*10^33)/(5.67*10^-8)(48330)^4

A=6.818*10^21 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(6.818*10^21/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

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