Question

2) The Plank radiation law is

I(lambda) = 2 pi h c²/[lambda⁵ (e ^{hc/labda kT} -1)]

take the derivative of this expression with respect to lambda to show

lambda_max = hc/(4.965kT).

You may uses 5 - x = 5 e^-x is true for x = 4.965.

Answer 1

The Planck radiation law is given by
  
The maximum of this function is calculated by taking derivative w.r.t. \lambda and is given by
   
  
  
Now the maximum of I implies
  
  
We now define
   
and so, the above transcendental equation becomes
   
  
  
   
And the solution of this equation is given by
   
  Using the definition of x, we get