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My understanding is that blackbody radiation occurs on a curve that depends on the temperature of the object - as the temperature increases, the area under the curve increases, and the peak of the curve moves to higher energies.

My understanding is that the curve is non-zero at every point, meaning that a photon of any energy has a chance of being emitted, albeit vanishingly tiny.

So, how would we calculate that chance for a given blackbody object and a given photon energy? I don't know what parameters matter, but I'd like to know for the general case and for the following specific case:

  • A blackbody object with temperature 0.1 K and surface area of 0.15 m^2 - roughly like a bowling ball.
  • A photon with frequency between 300 GHz and 430 THz - the infrared range, according to Wikipedia.
  • The amount of time one would have to wait to have a 95% chance of the photon being emitted.
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closed as off-topic by ACuriousMind, John Rennie, user36790, Kyle Kanos, Sebastian Riese Oct 1 '15 at 16:56

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  • $\begingroup$ You have to give a range of frequencies for this to be a well-formed question :) $\endgroup$ – DanielSank Oct 1 '15 at 0:26
  • $\begingroup$ Ah, thanks a bunch. I edited to just include the entire infrared range. $\endgroup$ – antiduh Oct 1 '15 at 0:31
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    $\begingroup$ Now that we have a well-formed question I have to give you the standard remarks. Welcome to Physics Stack Exchange. This is a site for physics question and answer. However, it's not a problem solving help site. We call questions looking for solutions to specific problems like this one "homework-like" and have some rules about whether or not they're considered on-topic. If you could please show your own work and ask about whatever issue has you stuck then you'll probably get an answer. For example, look up the blackbody radiation formula, integrate over a frequency range to get power... $\endgroup$ – DanielSank Oct 1 '15 at 0:41
  • $\begingroup$ ...and then convert that to photon flux. If you get stuck somewhere in that journey ask a specific question about that conceptual issue. $\endgroup$ – DanielSank Oct 1 '15 at 0:41
  • $\begingroup$ @DanielSank - Ah, like a newb, I hadn't read the help center first. I'll look up the formula and try to perform the calculation myself. If it's any help, I'm trying to refute some bad physics on reddit: reddit.com/r/askscience/comments/3myo92/…. Also, should I delete this question, given that it is not likely to be productive? $\endgroup$ – antiduh Oct 1 '15 at 0:48
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A few hints on how to do this calculation:

1) write down the equation for the total energy emitted per unit time at 0.1 K (Stefan-Boltzmann law).

2) write down the equation for the shape of the spectrum (Planck's Law)

Note that Planck's Law gives the energy in an interval: you would have to divide that by the energy of a photon at each wavelength in order to get the probability of an emission in unit time. Integrate (Planck)/$\hbar\lambda$ between the limits of wavelength of interest: this is the probability of an emission, $p$.

To get to 95% confidence that you have an emission, you need to consider that you have $(1-p)$ probability of no emission per unit time. This is a number that must get below 5%, so solve for $s$:

$$(1-p)^s = 0.05$$

Confirm by integrating (2) , over all wavelengths, and show you get (1).

You will want to do this using numerical integration. Pick a small enough step size, and be careful of rounding errors (numerical instability).

If you do the above and work it into your question (or write it as an answer yourself) we can help you figure out any missing parts without violating the "homework and exercises" policy.

UPDATE

I took a quick look at the magnitude of the numbers, and there is a problem...

Most of your math programs will not want to evaluate Planck's Law in the range of temperatures and wavelengths: the exponential term will overflow. You therefore have to make a simple approximation so you can evaluate the logarithm:

$$\begin{align}\\ B(\lambda, T) &= \frac{hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT}}-1}\\ &\approx\frac{hc^2}{\lambda^5e^{\frac{hc}{\lambda kT}}}\end{align}$$

when $\frac{hc}{\lambda kT} \gg 1$. Taking the natural logarithm, you find

$$\log(B) = \log(hc^2) - 5\log(\lambda) - \frac{hc}{\lambda kT}$$

I wrote a few lines of Python to evaluate this over the wavelength range of interest:

enter image description here

zooming in a lot more, you get this:

enter image description here

The probability of photon emission is very, very small indeed.

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