How Does Planck's Formula Change Without Stimulated Emission?

[UrbanXrisis]

I am to show Planck's blackbody radiation formula without the stimulated emission:

level n is the upper energy level, level m is the lower energy level:

N_m B u(\lambda,T)=N_n A

\frac{N_n}{N_m}=\frac{B u(\lambda,T)}{A}

N_n=ce^{-E_n /kT}

N_m=ce^{-E_m /kT}

\frac{N_n}{N_m}=e^{-h \nu / kT}

e^{-h \nu / kT}=\frac{B u(\lambda,T)}{A}

as T --> infinity

\frac{B u(\lambda,T)}{A}=1

u(\lambda,T)= \frac{A}{B e^{h \nu / kT}}

this is the formula for blackbody radiation if the stimulated emission was not placed into the equation right?

should I include that \frac{A}{B}=\frac{8 \pi hc}{ \lambda^5}?

I am also to find the range of wavelengths that is almost the same as Planck's formula... since the A/B for both formulas are the same... then i should find when \frac{1}{e^{hc/ \lambda kT}-1}= \frac{1}{e^{hc/ \lambda kT}} are aprox. equal?

I am also asked who had already discovered this formula? I have no clue, but my best guess is Rayleigh?

I am also asked which modern high tech device would not work if blackbody radiation was described by this new formula? my best guess would be the laser? since it needs stimulated emission to produce light.

how is my work for these problems?

 

[dextercioby]

A/B must be written in terms of the frequency of the electromagnetic radiation being subject to quantization. For small T, which means large values of the exponent of "e", the formula had been previously found by W.Wien.

Daniel.