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Q. 36

Expert-verifiedFound in: Page 1296

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

II In general, an atom can have both orbital angular momentum and spin angular momentum. The total angular momentum is defined to be $\overrightarrow{J}=\overrightarrow{L}+\overrightarrow{S}$. The total angular momentum is quantized in the same way as $\overrightarrow{L}$ and $\overrightarrow{S}$. That is, $J=\sqrt{j\left(j+1\right)h}$, where $j$ is the total angular momentum quantum number. The z- component of $\overrightarrow{J}$ is ${J}_{z}={L}_{z}+{S}_{z}={m}_{j}h$, where ${m}_{j}$ goes in integer steps from $-j$ to $+j$. Consider a hydrogen atom in a $p$ state, with $l=1$.

a. role="math" localid="1648549390842" ${L}_{z}$ has three possible values and ${S}_{z}$ has two. List all possible combinations of ${L}_{z}$ and ${S}_{z}$. For each, compute ${J}_{z}$ and determine the quantum number ${m}_{j}$. Put your results in a table.

b. The number of values of ${J}_{z}$ that you found in part a is too many to go with a single value of $j$. But you should be able to divide the values of ${J}_{z}$ into two groups that correspond to two values of $j$. What are the allowed values of $j$? Explain. In a classical atom, there would be no restrictions on how the two angular momenta $\overrightarrow{L}$ and $\overrightarrow{S}$ can combine. Quantum mechanics is different. You've now shown that there are only two allowed ways to add these two angular momenta.

According to the wave mechanics, the orbital quantum number / is restricted to the integer values from $0$ to $n-1$. That is, all the possible values of the orbital quantum number / depend upon the principal quantum number $n$. They are expressed as

$l=0,1,2,...n-1$

(a) $\frac{3}{2}$ and $\frac{1}{2}$

(b) Values of $j$ are $\frac{1}{2},\frac{3}{2}$

The magnetic quantum number $m$ is restricted to the integer values from $-/$ to $+/$. That is, all the possible values of the magnetic quantum number $m$ depend upon the orbital quantum number $l$. Mathematically, it is expressed as

$m=2l+1$

The spin quantum number ${m}_{s}$ correspond to different orientations of the axis of the spin of the electron. There are basically two values of the spin quantum number ${m}_{s}$, namely $-\frac{1}{2}$ and $+\frac{1}{2}$.

The magnitude of the spin angular quantum number obeys the quantization rule. Using the quantization condition, the equation for the magnitude of the spin angular momentum ${L}_{s}$ is given as,

${L}_{s}=mh$

Here, $h$ is Planck's constant and is equal to $h$ divided by $2p$.

For$l=1$, the magnetic quantum number varies as,

${m}_{1}=-1,0,1.$

By using the equation (1), the spin angular momentum varies as,

${L}_{2}=-1\left(h\right),0\left(h\right),1\left(h\right)\phantom{\rule{0ex}{0ex}}=-h,0,+h$

For each values of m, there are two $s$ states as,

$+\frac{1}{2}$ and $-\frac{1}{2}$

There are ${S}_{2}=\frac{1}{2}h$ and $-\frac{1}{2}h$ for each ${L}_{2}$.

The total angular momentum quantum number is,

${J}_{s}={L}_{2}+{S}_{2}$and ${J}_{z}={m}_{j}h$.

Now for ${L}_{2}=h$ we have two values of ${J}_{s}$ which are,

$h+\frac{1}{2}h=\frac{3}{2}h$ and $h-\frac{1}{2}h=\frac{1}{2}h$

Then the corresponding ${m}_{j}$ values are

$\frac{3}{2}$ and $\frac{1}{2}$.

${L}_{s}$ | ${S}_{z}$ | ${J}_{z}=({L}_{z}+{S}_{z})$ | ${m}_{j}=\frac{{J}_{z}}{h}$ |

$h$ | $\frac{1}{2}h$ | $\frac{3}{2}h$ | $\frac{3}{2}$ |

$h$ | $-\frac{1}{2}h$ | $\frac{1}{2}h$ | $\frac{1}{2}$ |

$0$ | $\frac{1}{2}h$ | $\frac{1}{2}h$ | $\frac{1}{2}$ |

$0$ | $-\frac{1}{2}h$ | $-\frac{1}{2}h$ | $\frac{1}{2}$ |

$$" width="9" height="19" role="math" style="max-width: none; vertical-align: -4px;" localid="1648552354339">$-h$ | $\frac{1}{2}h$ | $-\frac{1}{2}h$ | $-\frac{1}{2}$ |

$-h$ | $-\frac{1}{2}h$ | $-\frac{3}{2}h$ | $-\frac{3}{2}$ |

As $j=l+m$, and we have $l=1$ and $m=+\frac{1}{2}or-\frac{1}{2}$

The allowed values of $j$ are

$1+\frac{1}{2}=\frac{3}{2}$ and $1-\frac{1}{2}=\frac{1}{2}$

Allowed values of $j$ are $\frac{1}{2},\frac{3}{2}$.

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