How Does Planck's Law Simplify for High Frequencies?

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SUMMARY

Planck's Law, expressed as B(f) = [(2hf^3)/c^2]{1/[exp(hf/kT) - 1]}, simplifies under two conditions: when hf << kT and when hf >> kT. For the case of hf >> kT, the expression simplifies to B(f) = [(2hf^3)/c^2]exp(-hf/kT), where the exponential term dominates and the constant 1 in the denominator can be neglected. This leads to the Wien distribution, which describes blackbody radiation at high frequencies.

PREREQUISITES
  • Understanding of Planck's Law and blackbody radiation
  • Familiarity with Taylor series expansions
  • Knowledge of thermodynamic concepts, specifically temperature (T) and Boltzmann's constant (k)
  • Basic grasp of exponential functions and their behavior in limits
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  • Study the derivation of Planck's Law in detail
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Students of physics, particularly those focusing on thermodynamics and quantum mechanics, as well as researchers interested in blackbody radiation and its applications in modern physics.

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Can someone help me with this please?

B(f) = [(2hf^3)/c^2]{1/[exp (hf/kT) - 1]} - Planck law in terms of frequency.

"Derive expressions for B(f) for the cases i. hf << kT, ii. hf >> kT".

I've done the first bit, that's just using a Taylor expansion. The second bit is where I'm stuck. In my lecture notes, I have that for hf >> kT, B(f) = [(2hf^3)/c^2]exp(-hf/kT). What I don't understand is why it's like that. If hf >> kT, shouldn't that exponential tend to infinity?

Thanks.
 
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Yes,the exponential should much larger than 1 and therefore the 1 in the denominator would be neglected...The exp.would come on top with a "-" in its argument and from the Planck distribution,u'd be getting the Wien distribution...

Daniel.
 
OH! Thanks :smile:.
 

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