GR9277 #31


Problem


\prob{31}
In a state of the helium atom, the possible values of the total electronic angular momentum quantum number are
 0 only
 1 only
 0 and 1 only
 0, 1/2, and 1
 0, 1, and 2

Atomic}Spectroscopic Notations
Spectroscopic notation is given by , and it's actually quite useful when one is dealing with multiple particles. , respectively, for orbital angular momentum values of . for electrons. j is the total angular momentum.
Knowing the convention, one can plug in numbers to solve . Since the mainscript is a S, . The total angular momentum is .


Alternate Solutions 
eshaghoulian 20070930 14:03:16  Whoa. You have to be careful. Given and ,the possible values of are, in general, j=l+s, l+s1, l+s2,...,ls. Hund's rules tell you that can only equal or ls, depending on whether the subshell is more than halffilled. In this case, and ls are the same number, so it does not matter.  

Comments 
deneb 20181013 21:26:52  I\'m confused what the answer is describing. Is the s=1 referring to the entire helium atom? This particular helium atom has two electrons which each have s=1/2, so together s=1? And then since it\'s in the S orbital, l=0, then j=1?   Poop Loops 20081005 14:50:27  So is s = 1/2 or s = 1? Two different "s"? I understand S = 0 because of 0 orbital angular momentum.
But does s refer to spin or what?
gear3 20131009 05:48:01 
the spin of Helium is 1.That's it

  bucky0 20071101 14:09:11  to clarify, the correct answer is B   eshaghoulian 20070930 14:03:16  Whoa. You have to be careful. Given and ,the possible values of are, in general, j=l+s, l+s1, l+s2,...,ls. Hund's rules tell you that can only equal or ls, depending on whether the subshell is more than halffilled. In this case, and ls are the same number, so it does not matter.
Jeremy 20071104 15:44:38 
I don't think Hund's are involved in the solution. Hund's rules are only to determine the ground state, and problem isn't asking about that. I think the real point is that can be or any integer step above that, up to, and including, . In this case, , so is the only option.

DDO 20081104 19:35:23 
Jeremy that is Hund's 3rd rule.

iriomotejin 20101006 08:42:25 
Hund's rules deal with ground state, not with excited ones. The state of atom in this problem is obviously excited, so Hund's rules are irrelevant.

 

Post A Comment! 

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .

type this... 
to get... 
$\int_0^\infty$ 

$\partial$ 

$\Rightarrow$ 

$\ddot{x},\dot{x}$ 

$\sqrt{z}$ 

$\langle my \rangle$ 

$\left( abacadabra \right)_{me}$ 

$\vec{E}$ 

$\frac{a}{b}$ 





The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... 

