Probability per atom and per second for stimulated emission to occur

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The discussion focuses on calculating the probability per atom and second for stimulated emission in hydrogen plasma at 4500 ºC, using the given lifetime of the 2p state (1.6 ns). The initial approach involved calculating the decay constant A, yielding a value of 6.25 x 10^8 s^-1. Participants discussed the importance of radiation density and temperature, referencing Planck’s radiation law to relate these factors. A key equation for radiation balance was presented, linking the probabilities of stimulated emission and absorption. The conversation highlights the need for additional quantities like B and ΔE to complete the calculation, emphasizing the complexity of the problem.
Philip Land
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Homework Statement


We are investigating hydrogen in a plasma with the temperature 4500 ºC. Calculate the probability per atom and second for stimulated emission from 2p to 1s if the lifetime of 2p is 1.6 ns

Homework Equations


##A=\frac{1}{\Sigma \tau}##

$$A_{2,1} = \frac{8*\pi *h * f^3*B_{2,1}}{c^3}$$

The Attempt at a Solution


[/B]
hmmm, I'm not sure how to approach this problem. I took the inverse of the life time and got that A= ##6.25*10^8 S^{-1}.##

But I'm not sure where to start or what formulas to use.

The only formula I know of which takes temperature into account is
Doppler line width: ##\Delta F = constant * f_0 * \sqrt(T/M) ## which I can't see how to apply in this case at all.

Any input on where to start?
 
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The probability of stimulated emission will depend on the density of radiation around the atom, which depends on the temperature.
 
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mfb said:
The probability of stimulated emission will depend on the density of radiation around the atom, which depends on the temperature.
Thanks a lot! I manage to as you said find a relation between radiation density and temperature, (Planck’s radiation law).

Then I used a Radiation balance and solved for ##A_{21} = \rho (f) * \frac{N_1}{N_2}*B_{12}
-B{21}*\rho (f). ##

Where ##g_1*B_{12} = g_2*B_{21}## if we let g1=g2 we get ##B_{12}=B_{21}##

We also know from statistics that ##\frac{N_1}{N_2}= e^\frac{- \Delta E}{kT}##

But my question is. To get A (which I guess I'm supposed to get). I need B and ##\Delta E## But I don't have those quantities... (as I'm aware of).
 
Did you use the given lifetime already?
 
mfb said:
Did you use the given lifetime already?
Yes I used that to get the frequency, used in Plancks radiation law.
 
Philip Land said:
##A_{21} = \rho (f) * \frac{N_1}{N_2}*B_{12} -B{21}*\rho (f). ##
It might help to rearrange this equation as ##N_2A_{21} + N_2B_{21}\rho (f)= N_1B_{12}\rho (f) ##.
Interpret each of the terms. One of the terms is closely related to what you are asked to find.
 

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