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Q35E

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Found in: Page 281

### Modern Physics

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.

See the step by step solution

## Step 1: A concept:

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

## Step 2: (a) Values of the equation at  φ=0 and φ=2π :

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,

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

## Step 3: Conclusion:

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.

## Step 4: (b) simply that the second derivative of Φ(φ) is 0 :

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.