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How did Stefan Boltzmann find the constant to the Stefan Boltzmann law?

Answer
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Hint: First of all we need to know the Stefan Boltzmann law statement and from that we can define what we understand by Stefan Boltzmann’s Constant. Then we need to know Stefan Boltzmann law. And from that we can write equations and on further solving by integrating the equation and a few more steps. Finally we find the value of the Stefan Boltzmann constant.

Compete answer:
Stefan Boltzmann law:
According to this law the amount of radiation emitted from a body per unit time from a surface area of blackbody let say A at absolute temperature let say T is directly proportional to the fourth power of the temperature, T.
Now the total power that radiates from a body per unit area over all wavelengths of blackbody can be obtained by integrating Planck's radiation formula. Hence the radiated power per unit area as a function of wavelength can be represented as,
dPdλ1A=2πhc2λ5(ehcλkT1)
Where,
P is the power of radiation.
A is the surface area of the black body that radiates the light.
λ is the wavelength of emitted radiation.
h is Planck's constant.
c is the speed of light which is also a const.
k is the Boltzmann constant.
T is the absolute temperature.
On further simplifying the Stefan Boltzmann equation we will get,
d(PA)dλ=2πhc2λ5(ehcλkT1)
Now in integrating both side with respect to the wavelength λ and the limit is from 0 to we will get,
0d(PA)dλdλ=02πhc2λ5(ehcλkT1)dλ
Now on further solving we will get,
PA=2πhc20dλλ5(ehcλkT1)
Now substituting,
hcλkT=x(2)
Now differentiating x with respect to λ we will get,
hcλ2kTdλ=dx
Or, dλ=λ2kThcdx
Now rearranging equation (2) we will get,
h=xλkTc
Or, c=xλkTh
Now putting all these in equation (1) we will get,
PA=2π(xλkTc)(xλkTh)20(λ2kThc)dxλ5(ex1)
PA=2π(x3λ5k4T4h3c2)0dxλ5(ex1)
Cancelling and rearranging we will get,
PA=2π(k4T4h3c2)0x3dx(ex1)
Now we know that,
0x3dx(ex1)=π415
Hence,
PA=2π(k4T4h3c2)π415PA=(2k4π515h3c2)T4
Now we can write,
PA=σT4ε=σT4
Where,
σ=(2k4π515h3c2)
Putting all the constant value we will get,
σ=5.670×108Wm2K4

Note:
If a body us not a black body abortion then it will emit less radiation because and the equation is given as u=eσAT4 where, e is the emissivity which lies in between 0 to 1. Remember that c=3×108ms1, π=3.14, k=1.4×1023JK1 and h=6.626×1034m2kgs1.