91Ó°ÊÓ

In a hydrogen atom, the magnitude of orbital angular momentum is given by;

$L=\sqrt{l\left(l+1\right)}\sout{h}$

In a hydrogen atom, the z-component of orbital angular momentum is equal to;

$L_z=m_i\sout{h}$

The z-component of an electron's spin angular momentum is equal to;

$S_z=m_s\sout{h}$

The principal quantum number n=4 corresponds to the N shell.

From 0 to n-1, the orbital quantum number l can have n integer values.

The minimum and maximum values are $I_{min}=0$and $I_{max}=3$, respectively.

$\begin{array}{l}{L}_{min}=\sqrt{0\left(1\right)}h=0\\ L_{min}=0\end{array}$

Hence, the smallest orbital angular momentum $L_{min}=0$

localid="1664861800218" $\begin{array}{l}{L}_{max}=\sqrt{3\left(4\right)}\sout{h}=2\sqrt{3}\sout{h}\\ L_{max}=2\sqrt{3}(1.055×{10}^{-34}J.s)=3.65×{10}^{-34}J.s\\ L_{max}=2\sqrt{3}\sout{h}=3.65×{10}^{-34}J.s\end{array}$

Hence, the largest orbital angular momentum

$L_{max}=2\sqrt{3}\sout{h}=3.65×{10}^{-34}J.s$

As a result, the largest $m_l$equals the largest l, which in this case is $m_{I,max}=I_{max}=3$.

The z-direction was chosen as the orbital angular momentum's direction.

as a result of putting the values,

$$\begin{gathered} L_{z,max}=3\sout{h}=3\left(1.055×{10}^{-34}J.s\right)=3.16×{10}^{-34}J.s \\ {L}_{z,max}=3\sout{h}=3.16×{10}^{-34}J.s \end{gathered}$$

Hence, the largest orbital angular momentum the electron could have in any chosen direction is$L_{z,max}=3\sout{h}=3.16×{10}^{-34}J.s$

Hence, the largest spin angular momentum the electron could have in any chosen direction is$L_{z,max}=3\sout{h}=3.16×{10}^{-34}J.s$

$$\begin{gathered} \frac{S_z}{L_z}=\frac{\sout{h}/2}{3\sout{h}}=\frac{1}{6} \\ \frac{S_z}{L_z}=\frac{1}{6} \end{gathered}$$

Hence, the ratio of its spin angular momentum in the z-direction to its orbital angular momentum in the z-direction is 1:6.