Chapter 28, Problem 18P

To determine

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

Answer to Problem 18P

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 nitrogen $\left(Z=7\right)$

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:

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

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

Conclusion:

The required value is calculated.