Chapter 44, Problem 8SP

To determine

The four quantum numbers for a single hydrogen electron in its ground state.

Answer to Problem 8SP

Solution:

$1$, $0$, $0$, $+\frac{1}{2}$, or $-\frac{1}{2}$

Explanation of Solution

Introduction:

The values of the principle quantum numbers corresponding to the shells are,

$\begin{array}{cccccc}\text{Shells}=& K& L& M& N& O\mathrm{......}\\ n=& 1& 2& 3& 4& \mathrm{5......}\end{array}$

The values of the orbital quantum numbers corresponding to subshells are,

$\begin{array}{ccccc}\text{Subshells}=& s& p& d& f\mathrm{......}\\ l=& 0& 1& 2& \mathrm{3......}\end{array}$

The range of the orbital quantum numbers goes from $0$ to $\left(n-1\right)$.

The range of the magnetic quantum numbers $\left(m_l\right)$ goes from $-l$ to $0$ to $+l$.

The range of the spin quantum number $m_s$ is $\pm \frac{1}{2}$.

Explanation:

Understand that the ground state configuration of a hydrogen atom is $1s^1$.

Only one electron will be present in the shell of the hydrogen as the atomic number ofhydrogen is $1$.

Recall the values of the principle quantum numbers corresponding to the shells.

$\begin{array}{cccccc}\text{Shells}=& K& L& M& N& O\mathrm{......}\\ n=& 1& 2& 3& 4& \mathrm{5......}\end{array}$

Observe from the ground state configuration of a hydrogen atomthat the electron is present in the $1s$ subshell of the $K$ shell. So, the value of principle quantum number corresponding to $K$ shell is $1$. Thus, $n=1$.

Observe from the ground state configuration of a hydrogen atom that the electron is present in the $1s$ subshell.

Recall the values of the orbital quantum numbers corresponding to subshells are,

$\begin{array}{ccccc}\text{Subshells}=& s& p& d& f\mathrm{......}\\ l=& 0& 1& 2& \mathrm{3......}\end{array}$

Observe that for the $s$ subshell, the corresponding value of the orbital quantum number is $0$. Thus, $l=0$.

Write the magnetic quantum number corresponding to $l=0$.

$m_l=0$

Understand that the value of the orbital quantum number is $0$. So, $m_l=0$.

Write the value of spin quantum number.

$m_s=+\frac{1}{2}\text{ or }-\frac{1}{2}$

Understand that only one electron is present in the ground state of the hydrogen atom. So, it will be either spin up $\left(m_s=+\frac{1}{2}\right)$ or spin down $\left(m_s=-\frac{1}{2}\right)$. So, its spin quantum number will be $m_s=+\frac{1}{2}\text{ or }-\frac{1}{2}$.

Conclusion:

Hence, the four quantum numbers corresponding to the electrons present in the ground state of the hydrogen atom are $1$, $0$, $0$, $+\frac{1}{2}$, or $-\frac{1}{2}$.