A mathematical solution of the azimuthal equation (7-22) is $\mathbf{\Phi }\left(\phi \right)\mathbf{=}{\mathbf{Ae}}^{\sqrt{\mathbf{-}\mathbf{D\phi }}}\mathbf{+}{\mathbf{Be}}^{{}^{\sqrt{\mathbf{-}\mathbf{D\phi }}}}$ , which applies when $\mathbf{D}$ is negative, (a) Show that this simply cannot meet itself smoothly when it finishes a round trip about the z-axis. The simplest approach is to consider $\mathbf{\phi }\mathbf{=}\mathbf{0}$ and $\mathbf{\phi }\mathbf{=}\mathbf{2}\mathbf{\pi }$. (b) If $\mathbf{D}$ were $\mathbf{0}$, equation (7-22) would say simply that the second derivative $\mathbf{\Phi }\left(\phi \right)$of is $\mathbf{0}$. Argue than this too leads to physically unacceptable solution, except in the special case of $\mathbf{\Phi }\mathbf{\left(}\mathbf{\phi }\mathbf{\right)}$ being constant, which is covered by the ${\mathbf{m}}_{\mathbf{l}}\mathbf{=}\mathbf{0}$ , case of solutions (7-24).

Azimuthal quantum number specifies shape and angular momentum of the orbital.

Consider the given data as below.

$\mathrm{\Phi }\left(\mathrm{\phi }\right)={\mathrm{Ae}}^{\sqrt{-\mathrm{D\phi }}}+{\mathrm{Be}}^{{}^{\sqrt{-\mathrm{D\phi }}}}$ ….. (1)

Where, is the Azimuthal Angle, are the Arbitrary constants, and is the Azimuthal function.

$D=-\frac{1}{\Phi }\frac{{\partial }^2\Phi }{\partial {\phi }^2}$

Let, eq. (1) is continuous when $\phi =0$ and $\phi =2\pi$.

As you know that, for a function to be continuous at$\phi =0$ and$\phi =2\pi$, the value at$\phi =0$ andsh$\phi =2\pi$ould be equal.

Hence, for that to hold,

$$\begin{gathered} \Phi \left(0\right)=\Phi \left(2\pi \right) \\ A{e}^{\sqrt{-D}\times 0}+B{e}^{{-}^{\sqrt{-D}\times 0}}=A{e}^{{}^{\sqrt{-D}\times 2\pi }}+B{e}^{-\sqrt{-D}\times 2\pi } \\ A+B=A{e}^{{}^{\sqrt{-D}2\pi }}+B{e}^{-\sqrt{-D}2\pi } \end{gathered}$$ ….. (2)

Also, if its derivative is continuous

$$\begin{gathered} \sqrt{-D}\left(A-B\right)=\sqrt{-D}\left(A{e}^{\sqrt{-D}2\pi }+B{e}^{-\sqrt{-D}2\pi }\right) \\ A-B=A{e}^{\sqrt{-D}2\pi }+B{e}^{-\sqrt{-D}2\pi } \end{gathered}$$ ….. (3)

Now, by adding equation (2) and (3), you get,

$A=A{e}^{\sqrt{-D}2\pi }$ ….. (4)

Also, by subtracting equation (3) from (2), you get,

$B=B{e}^{-\sqrt{-D}2\pi }$ ….. (5)

For the equation (4) to hold,

Either $A=0$ or $D=0$

And for eq. (5) to hold

Either $B=0$ or $D=0$

You can’t have both$A=0$ and $B=0$, wave function will not be possible if it holds.

And if $D=0$, the given equation$\Phi \left(\phi \right)=A{e}^{\sqrt{-D}\phi }+B{e}^{{-}^{\sqrt{-D}\phi }}$will be a constant.

Hence, the given function cannot meet itself smoothly when it finishes a round trip about the z-axis.

If $D=0$, the equation $\Phi \left(\phi \right)=A{e}^{\sqrt{-D}\phi }+B{e}^{{-}^{\sqrt{-D}\phi }}$ will be a constant and will be a linear function, it will only repeat itself after $2\pi$ if the slope is zero.