Recent content by samjohnny

Homework Statement Homework Equations ##m = m_0 − 2.5 log_{10} f## where ##m## is the absolute magnitude of an object, and ##m_0## is the zero point. The Attempt at a Solution I'm having difficulty with is part b, the answer to which I believe is required for the subsequent parts. I...

So including all possible ##Js## gives ##J=0,1,2##?

Ok, then I'll implement LS coupling as found on this page. And so let ##L_1## be the orbital angular momentum of the ##5p## electron, and ##L_2## that of the ##6s## electron. Then ##L=L_1+L_2=1+0=1##. And similarly for the spins: ##S=\frac{1}{2}±\frac{1}{2}=1## or ##0##. Therefore, ##J=L+S=1##...

Ah OK, so both the ##5p## and ##6s## shells have contribution to the total angular momentum. And I went completely off track with ##l##. For the ##6s## orbital ##l=0## and ##s=\frac{1}{2}##. And so would that mean that ##j=s+l=\frac{1}{2}##, and so ##J=\sqrt{j(j+1)}=\frac{3}{4}ħ##? And for...

I see. Then the electron configuration of the first excited state would be: ##[Kr]4d^{10}5s^25p^16s^1##. Then for part b), since the total angular momentum of an atom is given by that of its outer electron, the total ##J## will be defined by the single outer electron in the ##6s## orbital...

Homework Statement An exercise examining the tin atom (Sn). Tin has a ground state electron configuration of ##[Kr]4d^{10}5s^25p^2##. a) Write down the electronic configuration of the first excited state. b) Illustrate with a vector diagram the allowed total angular momentum ##J## values for...

Ah right, where we use the fact that ##\left< E \right> = \left< N_R \right>E_R +\left< N_L \right>E_L ##. So, we've determined that L decreases as the stretched band is heated. So then ##LF## must also get smaller. However, the fact that there is a minus sign (##\left< E \right> = -LF##)...

Ah right, since we have the probability of a link pointing to the right, and we can yield the same for a link pointing to the left, we can simply compute the expectation/average value as ##\left< N_R \right> = NP_R## where ##P_R## is the result of part b). Doing the same for the ##\left< N_L...

Homework Statement Homework Equations $$ Z(1) = \sum_{i=1}^{} e^{\frac{E_i}{K_bT}} $$ where ##E_i## is each of the possible energy states available to a single link (in this case the right and the left states). $$ P=\frac{\sum_{i=1}^{} e^{\frac{E_i}{K_bT}}}{Z} $$ The Attempt at a Solution...

Conventions always throw me! Thanks a lot for the help.

Ah yes, total internal reflection. Thank you. Also, this seems trivial but I'm now having trouble obtaining the angle ##x##. I've attached my working.

That makes sense; I set the condition for when the reflection coefficient is zero but failed to follow it through to the final condition for the thetas. As for the second part of the question, by geometry we have that the angle of incidence at the second interface is ## x = \theta_2 - \gamma...

Ok here I'm using a hunch that ##\theta_1+\theta_2=\frac{\pi}{2}## . Doing so, and applying snells law at the bottom interface gives: $$ n_g\sin\theta_2 = n_a\sin\theta_1 $$ But ##\theta_1=-\theta_2+\frac{\pi}{2}## Substituting this into the equation by Snells law gives: $$ n_g\sin\theta_2...

Anyone? :)

Homework Statement Homework Equations ## θ_{brewster} = arctan(\frac{n_2}{n_1}) ## The Attempt at a Solution Hi all I'm having difficulty on this part of a question Now, showing that all light passes through is to say that there be no reflection. This occurs when the angle of incidence...