What Wavelength Maximizes u(lambda) for a 50,000 K Star?

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SUMMARY

The maximum wavelength for energy density, u(lambda), of a star with a surface temperature of 50,000 K can be determined using Planck's law. The formula for u(lambda) is given by u(lambda) = 8(pi)hc/(lambda^5*(e^(hc/kTlambda)-1). To find the wavelength at which u(lambda) is maximized, one must differentiate this function and set the derivative to zero. The discussion emphasizes the use of the product rule for differentiation to simplify the process of finding the maximum energy density.

PREREQUISITES
  • Understanding of Planck's law
  • Familiarity with differentiation techniques, specifically the product rule
  • Knowledge of thermodynamic concepts related to blackbody radiation
  • Basic algebra and calculus skills
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  • Study the application of Planck's law in astrophysics
  • Learn advanced differentiation techniques in calculus
  • Explore the concept of blackbody radiation and its significance in stellar physics
  • Investigate the relationship between temperature and wavelength in Wien's displacement law
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Astronomy students, physicists, and anyone studying stellar properties and blackbody radiation will benefit from this discussion.

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Homework Statement


At what wavelength is u(lambda) a maximum for a star with a surface temperature of 50,000 K?


Homework Equations


Planck's law
u(lambda)=8(pi)hc/(lambda^5*(e^(hc/kTlambda)-1)

The Attempt at a Solution


I think the maximum is where the derivative of the function u(lambda)=0, but the derivative is too messy to solve in terms of lambda so I can't find the wavelength. So I think I'm way off as far as how to solve it. Any help would be appreciated.
 
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well, the differentiation should be done in order to get \lambda at maximum energy density. The differentiation isn't too hard, try using the product rule first.
 

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