Chapter 28, Problem 17P

(a)

To determine

The quantum numbers for each electron in the ground state of carbon.

Answer to Problem 17P

The answer is $\left(1,0,0,-\frac{1}{2}\right),\left(1,0,0,+\frac{1}{2}\right),\left(2,0,0,-\frac{1}{2}\right),\left(2,0,0,+\frac{1}{2}\right),\left(2,1,-1,-\frac{1}{2}\right),\left(2,1,-1,+\frac{1}{2}\right)$

Explanation of Solution

Given info: Element carbon $\left(Z=6\right)$

Formula used:

The principal quantum number $n$ can take on any integer value $\left(1,2,3,\mathrm{...}\right)$ and also, corresponds to the quantum number of the old Bohr theory. The orbital quantum number $l$ can take on values from $0$ to $\text{n-1}$ .

The magnetic quantum number, $m_l$ , can take on the integer values from $-l$ to $+l$

The spin quantum number, $m_s$ , can be $+\frac{1}{2}$ to $-\frac{1}{2}$

Calculation:

Now, start with $n=1$ shell, and also, list quantum numbers in order $\left(n,l,m_l,m_s\right)$ ;

  $\begin{array}{l}\left(1,0,0,-\frac{1}{2}\right),\left(1,0,0,+\frac{1}{2}\right),\\ \left(2,0,0,-\frac{1}{2}\right),\left(2,0,0,+\frac{1}{2}\right),\\ \left(2,1,-1,-\frac{1}{2}\right),\left(2,1,-1,+\frac{1}{2}\right)\end{array}$

(b)

To determine

The quantum numbers for each electron in the ground state of magnesium.

Answer to Problem 17P

The answer is $\begin{array}{l}\left(1,0,0,-\frac{1}{2}\right),\left(1,0,0,+\frac{1}{2}\right),\left(2,0,0,-\frac{1}{2}\right),\left(2,0,0,+\frac{1}{2}\right),\left(2,1,-1,-\frac{1}{2}\right),\left(2,1,-1,+\frac{1}{2}\right),\\ \left(2,1,0,-\frac{1}{2}\right),\left(2,1,0,+\frac{1}{2}\right),\left(2,1,-1,-\frac{1}{2}\right),\left(2,1,-1,+\frac{1}{2}\right),\left(3,0,0,-\frac{1}{2}\right),\left(3,0,0,+\frac{1}{2}\right)\end{array}$

Explanation of Solution

Given info: Element magnesium $\left(Z=12\right)$

Formula used:

The principal quantum number $n$ can take on any integer value $\left(1,2,3,\mathrm{...}\right)$ and also, corresponds to the quantum number of the old Bohr theory. The orbital quantum number $l$ can take on values from $0$ to $\text{n-1}$ .

The magnetic quantum number, $m_l$ , can take on the integer values from $-l$ to $+l$

The spin quantum number, $m_s$ , can be $+\frac{1}{2}$ to $-\frac{1}{2}$

Calculation:

Note that, without additional information, there are other possibilities for the last three electrons.

So, the quantum numbers for each electron in the ground state of magnesium:

  $\begin{array}{l}\left(1,0,0,-\frac{1}{2}\right),\left(1,0,0,+\frac{1}{2}\right),\left(2,0,0,-\frac{1}{2}\right),\left(2,0,0,+\frac{1}{2}\right),\left(2,1,-1,-\frac{1}{2}\right),\left(2,1,-1,+\frac{1}{2}\right),\\ \left(2,1,0,-\frac{1}{2}\right),\left(2,1,0,+\frac{1}{2}\right),\left(2,1,-1,-\frac{1}{2}\right),\left(2,1,-1,+\frac{1}{2}\right),\left(3,0,0,-\frac{1}{2}\right),\left(3,0,0,+\frac{1}{2}\right)\end{array}$