2) The Plank radiation law is I(lambda) = 2 pi h c2/[lambda5 (e hc/labda kT -1)] take the...

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Physics

2) The Plank radiation law is

I(lambda) = 2 pi h c2/[lambda5 (ehc/labda kT -1)]

take the derivative of this expression with respect to lambda toshow

lambdamax = hc/(4.965kT).

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

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The Planck radiation law is given by The maximum of this    See Answer
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