I was studying this chapter lately, and this part of chapter always confuse me. I often think why $3d$ orbital in transition metal is higher in energy than 4s orbital? (the whole question will refer to transition elements). My confusion arises when I look at the explanation on a website (attached below), it says that because ionization energy always removes the electron from the highest, which mean that first removed electron is the last electron being put in the orbital, so I conclude that $4s$ is higher than 3d in transition element. But I found another website (attached below) stating that the highest orbital is $3d$. Now my question which one is the correct one? First link:  Second link:
I'm currently taking Quantum Mechanics in school and I came across a table displaying the following in the given order:   $E_n (\text{energy levels}), g_n (\text{eigen functions})$   $n \ l, [m]$  $E_1, 1$   $1s, [0]$  $E_2, 4$   $2s, [0]$ $2p, [-1,0,1]$  $E_3, 9$   $3s, [0]$ $3p, [-1,0,1]$ $3d, [-2,-1,0,1,2]$    Now, as I understand $n$ defines the size of the orbital, $l$ the shape and $m$ the orientation.  My question is, how does the atoms shell model fit into this if one can have two electrons per orbital? Are $1s$ the same as the orbital $2s$, but just renamed for the sake of higher energy levels? Or do two similarly looking orbits exist for different energy levels? For example two "layers" of spherical electron clouds for an atom with $4$ electrons.  I just can't see how $2$, $8$, $18$ becomes $2$, $8$, $8$ as we were taught in Physics at high school? Am I missing something?
Shells have a useful feature where pressing the keyboard UP arrow replaces the contents of the current command line with the previous command that was entered. (And if you do it again, the command before that, and so on.)  I would like a similar feature, where pressing (some other button) replaces the contents of the current command line with the last line of output from the previous command (and if you do it again, the previous line of output , and so on.)  Is there any shell that provides a feature like this?  The motivation is that you often want to run a command, and then run another command with part of the previous commands's output as input. (No, I don't want to do anything involving piping the output of one command into the next. I'm looking for an interactive experience, where you run the first command, see its output, and then get a line of that output into your next command, without having to reach for the mouse to do a clumsy copy-and-paste.) Moving /var, /home to separate partition I am attempting to move some folders (such as /var and /home) to a separate partition after reading this guide:    I was able to move one folder successfully following guide.  However, it doesn't seem to work for multiple folders, and all my folders are dumped into the partition without proper folders.  I would like to mount /var, /home, and /tmp onto the separate partition; can someone guide me on this? Rings and some of their properties So I am studying for an exam and was wondering if anyone could explain some of the reasoning being the proofs of the following properties:   Let R be a ring with $a,b\in R$. Then    $a0=0a=0$  $a(-b)=(-a)b=-ab$ $(-a)(-b)=ab$   Proof 1:   $a0=a(0+0)=a0+a0$ hence $a0=0$   I don't really understand why $a0+a0$ proves that $a0=0$ LM317 getting hot Should an LM317 (specifically a ) rated for 500ma output current really be getting hot to touch when only pulling 170ma at 5V?  Package is SOT-223, input is 9V.  Is this normal? ^x^y unix trick for all instances in last command? One of my favorite Unix tricks is ^x^y, which will take the last command and replace the first instance of "x" with "y". However, I'm wondering if a similar trick works to replace all instances of "x" with "y" in the last command? Is there an "exists" function for jQuery? How can I check the existence of an element in jQuery?  The current code that I have is this:  if ($(selector).length &gt; 0) {     // Do something }   Is there a more elegant way to approach this? Perhaps a plugin or a function? "I", "me" and "myself" Possible Duplicate:           What is correct? We are a family of four: my father, my mother, my brother and me. or We are a family of four: my father, my mother, my brother and I. or We are a family of four: my father, my mother, my brother and myself. Cannot delete files, nothing seems to work I am on Windows 7.  I have some files on my computer leftover from an old version of Photoshop.  (I think... I don't remember exactly where they came from, but they seem to be Adobe files.)  There are 124 of them in a folder, which I have named "KILL."  (I don't remember what the original name of this folder is, it's been on my comptuer a long time.)  These files seem to be absolutely undeletable, as far as I can tell.  I can move them around and change the name of the folder, but deleting them is out of the question.  I also can't move them onto a different hard drive or flash drive, I've tried that.  So, let me list the things I've tried first:   A whole bunch of different "unlocking programs."  Including: IObit Unlocker, Lock Hunter, FileAssassin, and Project Kill box.  What I use LockHunter I am told: "No processes locking this file or folder have been found."  All of the other programs I've tried give me similar messages, telling me that nothing is locking the files.  Yet when I try to delete the files using LockHunter, I am simply told "Cannot delete x."  No reason is given. Tried using cmd.exe.  Even when I run a command window under administrator, it won't even let me access the KILL folder, much less delete anything.  I also tried killing explorer.exe from the task manager, and then using the cmd window.  Still can't even access that folder. Tried using safe mode in combination with these other options, doesn't change a thing. Tried changing permissions of the files using the security tab under properties.  It simply says "not inherited."  It will not allow me to change a thing. I am definitely the administrator of this computer. Tried Disk Cleanup and Defrag.  I suspected that some files must be damaged or something, but Disk Cleanup wasn't able to find any errors.  Defrag didn't help.   So, let me just say that I am not a computer expert.  But based on all the things I've tried, I suspect that this might be some hardware problem?  It seems strange that these files don't seem to be used by any processes, and yet are absolutely undeletable no matter what I do.  Is there anything else I can try?  Has anyone had any similar experience, and have any idea why this might have happened?  EDIT: Interesting recent development.  As I have toyed with moving the folder around, I moved it to the Program Files folder.  Now that it's there, it suddenly CANNOT be moved nor renamed.  It is stuck there, apparently permanently, as a folder called "KILL."  How irritating. I'd like to show that the class of all limit ordinals, LIM, is cofinal in the class of all ordinals ON.  I thought that I could do this by contradiction and assume that it wasn't cofinal in ON. Then there is an $\alpha \in$ ON such that for all $\beta \in $ LIM: $\beta &lt; \alpha$, i.e. $\beta \in \alpha$ and hence LIM $\subset \alpha$.  This means that LIM is a set so we can consider $\bigcup$ LIM. Then we have $\beta &lt; \bigcup$ LIM for all $\beta \in$ LIM by definition of $\bigcup$ LIM. So LIM $\subseteq \bigcup$ LIM.   $\bigcup$ LIM is either a limit or a successor ordinal.  If it is a limit ordinal then $\bigcup$ LIM $\in$ LIM and by transitivity of ON, $\bigcup$ LIM $\subseteq$ LIM, so $LIM = \bigcup$ LIM in this case which would mean that $\bigcup$ LIM $\in$ LIM $= \bigcup$ LIM which would be a contradiction because sets cannot contain themselves. Hence $\bigcup$ LIM cannot be a limit ordinal.  If $\bigcup$ LIM is a successor ordinal then $\bigcup$ LIM = $\beta \cup \{&#160;\beta \}$ for some $\beta \in $ ON.  Now I'm not sure how to proceed from here. I'm also not sure about whether I'm on the right track with this proof because I think there should be a shorter way  to prove this.  Thanks for your help. creating variable using variable value as part of new variable name I'm trying to create a new variable using the value of an existing variable as part of the variable name.  filemsg"$word1"=" "   I've also tried   filemsg$word1=" "  filemsg${word1}=" "   on all attempts I get the following when that line executes,  cicserrors.sh[45]: filemsgCICS= : not found [No such file or directory] I'm writing a script and I would like to pass the results from bc into a variable. I've declared 2 variables (var1 and var2) and have given them values.  In my script I want to pass the results from bc into another variable say var3 so that I can work with var3 for other calculations.  So far I have been able write the result to a file which is not what I'm looking for and also I've been able to echo the result in the terminal but I just want to pass the result to a variable at moment so that I can work with that variable.  echo "scale=2;$var1/var2" | bc I was just reading Harry Potter and the Deathly Hallows, and I was wondering, when Hermione Granger changed into Bellatrix Lestrange, she should've obtained the Dark Mark, temporarily.  Now, my question is: would the Dark Mark work if pressed? I followed Mike's solution from my first question:    My next question is how to add entries (for the abstract and acknowledgements pages) into the table of contents. "Effect" vs. "Affect" I've noticed that some people use effect and affect interchangeably. What are the differences between these two and when are the proper situations to use each of them? With constant travel, is it possible to not have a tax residence? Find $\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right]   \end{align*}$.   After some minutes around this limit I did it this way:  $\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1$  So,$\space x=2^y+1$.  When $x \to +\infty$,$\space y \to +\infty$ also. By substitution:  $\begin{align*} \lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y+1-1)}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y)}{2^y+1}\right]=\end{align*}$  $\begin{align*}\lim_ {y \to+\infty} \left [ \frac{y}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y+1}{y}}  \right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y}{y}+\frac{1}{y}}\right]= \frac{1}{+\infty+0}=0  \end{align*}$  Is this correct?Are there any other easy way to find this limit?Thanks What is the value of $1^i$? $\,$ Wilson and Fermat's Theorem Question!     Show that the pair of positive integers $n$ and $n+2$ are twin primes if and only if $$4[(n-1)! + 1] + n = 0 \ (\text{mod} \ n(n+2)),$$ where $n$ does not equal $1$. An $n \times n$ matrix $M$ of integers has the property that $M_{ab} = M_{xy}$ iff $a+b=x+y$. How many distinct elements are there in $M$? Radio antenna producing waves in the visible spectrum If a radio could produce waves in the visible light spectrum, what would the result be?  This is a thought experiment that I've pondered for a few years now.  I realize there are a few/many real-world constraints, but if we lifted these constraints for the sake of thought, what could we expect?  Personally, I don't see why we wouldn't observe visible light emitting from the antenna, disregarding any light from Blackbody Radiation.