(Solved): 38. Deriving Planck Blackbody Spectrum continued... 8. Exploration: Deriving the Planck Bla ...

38. Deriving Planck Blackbody Spectrum continued... 8. Exploration: Deriving the Planck Blackbody- (b) Use the equation from Part (a) to solve for the constant C. (This is called "normalizing" Spectrum the probability distribution.) You will need to In this problem you are going to derive the average know that the sum of an infinite geometric energy $E_w(?)$ and blackbody spectrum $S(?)$, based series with first term $a_1$ and common ratio on Planck's hypothesis: a single wave of frequency $r<1$ is $a_1/(1?r)$. $v$ must have energy $nhv$, where $n$ can be $0,1,2$, etc. (c) To find the average you evaluate the sum of but nothing in between. each possible energy times its probability: (a) If you measure a particular wave of frequency $v$ at a particular moment, the probability that you will find energy $E$ is proportional to $e^{?E/\left(k_{B}T\right)}$. Using $C$ for the constant of proportionality, we can say that the measurement (Take a moment to convince yourself that this $E=nh?$ has probability $C{e}^{?nh?/\left(k_{B}T\right)}$. Explain is the same calculation as we did in the text, briefly why the following equation must be although we never quite wrote it in this way.) Plug in the value of $C$ that you found in Part (b) true: and then evaluate the sum. ${?}_{n=0}^{?}C{e}^{?nhv/\left(k_{B}T\right)}=1$ (d) Plug the formula you found for $E_w$ into Equation (3.4) to find the Planck spectrum. Part (c) refers to this problem from the text. Example: Calculating the Average Energy with Only Two Options Suppose a wave in our cavity could only have two possible energies: $E=0$ and $E=2k_{B}T$. you measured its energy many times, what would be the average of all your measurements? Like our previous example, this "only two possible energies" scenario is entirely hypothetical But if you can follow the probability work in this example, then you can replicate many of Planck's calculations. Answer: You might suppose that the average would be $k_{B}T$ (the average of the two possible measurements), but that doesn't take into account the probabilities. Equation (3.7) says that lower energy levels are more likely than higher energy levels, so you will measure 0 far more often than $2k_{B}T$. The average must still be between 0 and $2k_{B}T$, but it should come out considerably lower than $k_{B}T$. Now let's make that quantitative. Equation (3.7) gives a proportionality relationship. We can express that relationship as an equation by introducing a constant of proportionality, which we will call C. So here are the probabilities of our only two possible measurements: $\begin{array}{l}P(0)=C{e}^{?0/\left(k_{B}T\right)}=C\\ P\left(2k_{B}T\right)=C{e}^{?\left(2k_{B}T\right)/\left(k_{B}T\right)}=C^{?2}\end{array}$ Now, imagine that you measure the energy of this wave exactly $\mathrm{C}+{\mathrm{Ce}}^{?2}$ times. Based on the above probabilities, you would expect $\mathrm{C}$ of your measurements to get an energy of 0 , and ${\mathrm{Ce}}^{?2}$ of your measurements to get an energy of $2k_{B}T$. (You might object that this is only possible if $\mathrm{C}+{\mathrm{Ce}}^{?2}$ happens to be an integen - you can't make half a measurement - but that's irrelevant for the following calculation.) The average energy you measure is the total of all your measured energies, divided by the number of measurements: $E_w=\frac{(0)(C)+\left(C^{?2}\right)\left(2k_{B}T\right)}{C+C{e}^{?2}}?0.24k_{B}T$ This calculation is called finding the "expectation value" based on a probability distribution. We did not just determine that the average energy of a wave is $0.24k_{B}T$. It isn't, because we just made up this example. Rather, we showed that this would be the average energy of a completely hypothetical wave for which 0 and $2k_{B}T$ are the only possible energies. $S(v)=\frac{8?v^2}{c^3}{E}_w(v)\text{ Book Eq. (3.4) }$