1. There are just 3x 1027 it air molecules in an otherwise empty room. Calculate
(a) the average number that will be in the front third of the room at any time,
(b) the standard deviation about this value,
(c) the relative fluctuation.
2. For air at temperature the probability that any one molecule is in an excited electronic state is about 10-10 (p = 10-1( q≈1). In a typical room there are about 1028 air molecules.
For this case, calculate
(a) the mean number of excited molecules,
(b) the standard deviation,
(c) the relative fluctuation.
3. In the derivation of the Gaussian form of the probability distribution, PN(n), we showed, using ln ab = ln a+ ln b, ln a/b = ln a - ln b, ln ab = b ln a, that
ln P(n) = ln N!/n!(N - n)!pnqN-n.
= ln N! - ln n! - ln(N - n)! + nlnp + (N-n)lnq.
(a) Rewrite this expression, expanding all the factorials on the right-hand side using Stirling's formula,
ln m! ≈ m ln m - m + 1/2 ln2Πm.
(b) Show that
ln P(n = n-) = 1/2.ln1/2ΠNpq = 1/2. ln1/2Πσ2
using n- = Np, q =1- p, ln ab = lna + ln b.
(c) Take the derivative of the expression for in P(n) in part (a) and drop terms that go to zero as n gets very large. Then evaluate this at n = n- = Np to show that d/dnln P(n)|n=n- = 0.
(d) Following a procedure like that in part (c), take the second derivative and show that d2/dn2 ln P(n)|n=n-= -1/Npq = -1/σ2
(e) Combine these results and the Taylor series expansion to show that P(n) = 1/√2Πσe-(n-n-)2/2σ2.
You are interested in the number of heads when flipping 100 coins. In the Gaussian approach, with PN(n) = Ae-B(n-n-)2, what are the values of the constants A and B? Find the probability of obtaining exactly the following numbers of heads: (a) 50, (b) 48, (c) 45, (d) 40, (e) 36.
The Gaussian distribution that we derived is ofthe form PN(n) = Ae-B(n-n-)2 where A and B are constants. Suppose that we have for the first derivative in the Taylor series expansion
d/dn lnP(n)|n = n- = ε,
where ε is small but not zero. What would the corresponding form of PN(n) be in this case?