Recent content by wc2351

Ah yes, I should have posted the actual (wrong) numbers in the first place. I'll behave better next time. Thanks for pointing out the ideal gas law; 1 atm * 1 / (k*300K) gives me 2.45 X 10^(25) particles per cubic meter, which is pretty close to the value I found above.

Well you see my work above, a differential equation. Anyway I found out that it was simply an error of plugging in wrong numbers. Here is the answer I got. total cross section = 5. 1 x 10 ^(-31) m density of nitrogen = 1.251 kg / m^(3) (I don't know the source) / (mass of nitrogen molecule) ~...

I was reading Wikipedia article on Rayleigh scattering and came upon this: "...the major constituent of the atmosphere, nitrogen, has a Rayleigh cross section of 5.1×10^(−31) m^2 at a wavelength of 532 nm (green light). This means that at atmospheric pressure, about a fraction 10^(−5) of...

Homework Statement The problem is from Ashcroft&Mermin, Ch32, #2(a). (This is for self-study, not coursework.) The mean energy of a two-electron system with Hamiltonian \mathcal{H} = -\frac{\hbar^2}{2m}(∇_1^2 + ∇_2^2) + V(r_1, r_2) in the state ψ can be written (after an integration by...

Thank you all for your detailed replies. I'll try to check out some of the suggested references once my final exams are over.

Hi, It never occurred to me before, but when you derive the spin-orbit Hamiltonian as a perturbation to the hydrogen Hamiltonian, you imagine your electron orbiting around the nucleus, and of course that's not the correct picture because of the electromagnetic radiation that leaks out. So I...