GREPhysics.NET: GR8677 #30

GR8677 #30

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Atomic $\Rightarrow$ }Orbitals

Recall that:

$\begin{eqnarray} s&=&0\\ p&=&1\\ d&=&2\\ f&=&3\\ g&=&4...(etc.),<br /> \end{eqnarray}$

where the right side of the equal sign is the angular momentum $l$ .

And then there's also the formula $2(2l+1)$ , which determines the number of electrons in each subshell.

(A) The configuration does not specify anything about $3d$ . This is the first row of transitional metals.

(B) From the formula above, one sees that the $s$ subshell needs $2$ electrons to be filled. The $4s$ subshell is thus not filled.

(C) $l=4$ corresponds to $g$ . The configuration does not specify $g$ .

(D) Because the problem states that potassium is in the ground state, the atomic number is the same as the number of electrons in the configuration. The sum of the superscripts is 19.

(E) The valance shell is $4s^2$ . Many-electron atoms have wave functions that take after the hydrogen atom, and one recalls that $\psi_{40m}$ looks like a spherically symmetrical blob of a ball. This is it.

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Comments

kyros
2007-10-30 21:42:24

Shouldn't that be Sigma_4,0,0, since m can only range from -l to +l anyway?

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simpsoxe
2006-11-30 21:20:14

well the valence shell is $4s^1$ , but my question is about the other electrons in the lower shells. don't we have to consider them as part of the electron distribution? or do we assume they all have spherically symmetric distributions? (and why would that be?)

Blake7
2007-07-23 18:26:38

The lower shells all appear to be completely filled

2 2 6 2 6

if I'm not mistaken that will make them all symmetric

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Shouldn't that be Sigma_4,0,0, since m can only range from -l to +l anyway?

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