91影视

Angular orbital momentum has value,$m_2=卤3$

By using the concept of quantum numbers, we can find the number of values and greatest allowed values.

Formulas:

Number of different values of$m_1$is given byrole="math" localid="1663138105073" $\left(2l+1\right)$

$L_{orb,z}=\frac{m_{l}h}{2蟺}$

For the given value of $l,m$varies from $-lto+l.$ Thus, in this case, $1=3$, and the number of different values of $m_1$is

$2l+1=2\left(3\right)+1=7$

So, there are 7 different values of $L_{orb,z}$

Similarly, since ${渭}_{orb,z}$is directly proportional to$m_1$, there are total 7 different values of ${渭}_{orb,z}$.

As we know,$L_{orb,z}=\frac{m_{l}h}{2蟺}$. So, the greatest allowed value of$L_{orb,z}$is given by

$\frac{{\left|m_l\right|}_{maximum}h}{2蟺}=\frac{3h}{2蟺}$

Greatest allowed magnitude of $L_{orb,z}$islocalid="1663139848998" $\frac{3h}{2蟺}$

Since${渭}_{orb,z}=-m_l渭B$, the greatest allowed value of${渭}_{orb,z}$is given by

${\left|m_l\right|}_{maximum}渭B=\frac{3eh}{4蟺m_e}$

Greatest allowed magnitude of${渭}_{orb,z}$ is$\frac{3eh}{4蟺m_e}$

The z component of the net angular momentum of the electron is given by

$$\begin{gathered} L_{net,z}=L_{orb,Z}+L_{s,Z} \\ =\frac{m_{l}h}{2蟺}+\frac{m_{s}h}{2蟺} \end{gathered}$$

From the given value,$m_l=3\&m_s=\frac{1}{2}.$ Thus,

role="math" localid="1663140942371" $$\begin{gathered} {\left[L_{net,z}\right]}_{maximum}=\frac{\left(3+\frac{1}{2}\right)h}{2蟺} \\ =\frac{3.5h}{2蟺} \end{gathered}$$

Greatest allowed magnitude for z component of the electron鈥檚 net angular momentum is$\frac{3.5h}{2蟺}$

Since, the values of$L_{\left(net,z\right)}$ are given by$\frac{{\left[m_j\right]}_{maximum}h}{2蟺}$with${\left[m_j\right]}_{maximum}=3.5$

The number of allowed values for z component of$L_{net,z}$ is given as

role="math" localid="1663141133934" $$\begin{gathered} 2{\left[m_j\right]}_{maximum}=2\left(3.5\right)+1 \\ =8 \end{gathered}$$

Number of values allowed to magnitude for z component of the electron鈥檚 net angular momentum is 8.