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Q35E

Expert-verifiedFound in: Page 281

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**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).**

(a) The given function cannot meet itself smoothly when it finishes a round trip about the z-axis.

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

**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 ** $\phi =0$** and sh$\phi =2\pi $ ould be equal.

Hence, for that to hold,

** **

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

Also, if its derivative is continuous

$\sqrt{-D}\left(A-B\right)=\sqrt{-D}\left(A{e}^{\sqrt{-D}2\pi}+B{e}^{-\sqrt{-D}2\pi}\right)\phantom{\rule{0ex}{0ex}}A-B=A{e}^{\sqrt{-D}2\pi}+B{e}^{-\sqrt{-D}2\pi}$ ….. (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.

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