Find the power emitted (quantium mech)

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To find the power emitted by a blackbody at 7000K through a 1m hole between wavelengths of 4000 and 4001 angstroms, the relevant equation involves the power spectral density. While one approach involves integrating over the specified wavelength range, the solution provided simplifies the calculation by substituting 4000 for λ and using a small dλ of 10^-10. This simplification is acceptable due to the minimal change in the expression over such a small range. The discussion confirms that both methods yield valid results, with the integration being more precise. Ultimately, the choice of method depends on the context and required accuracy.
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Homework Statement


consider a blackbody at t=7000 what is the power emitted through a hole 1m between\lambda= 4000 and 4001 angstrom


Homework Equations



P(\lambda) d\lambda=A(2\pihc2/\lambda5(...)d\lambda

The Attempt at a Solution


my question is the math, I thought u had to make a integral on both sides and have it go from 4000 to 4001, but the solution does not do this. they just plug in 4000 for \lambda and for d\lambda they have 10-10 Is both ways correct? I don't remember what i got when i did the integral
thanks
 
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Strictly speaking, your way is "more correct". They use the fact that the range for lambda is very small and the expression is not going to change a lot. This way, the integration turns into a multiplication.
 
ok thanks
 
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