91Ó°ÊÓ

The type of scattering potential is known as hard-space scattering.

Consider scattering by a hard sphere, for which the potential is infinite for$r<a$, and zero for$r>a$.

It follows that $\mathrm{Ψ}\left(r\right)$ is zero in the region $r<a$, which implies that ul = 0 for all l .

From equation 11.33, the's for a hard sphere is given by:

$a_l=i\frac{j_l\left(ka\right)}{k{h}_l^{\left(1\right)}\left(ka\right)}$ ……. (1)

The relation between and is given by equation as:

$a_l=\frac{1}{k}{e}^{i{\mathrm{δ}}_l}\sin \left({\mathrm{δ}}_l\right)$ …… .. (2)

Combine these two equations we get:

$e^{i{\mathrm{δ}}_l}\sin {\mathrm{δ}}_l=i\frac{j_l\left(ka\right)}{h_l^{\left(1\right)}\left(ka\right)}$

From the definition of $h_l^{\left(1\right)}$we have:

$h_l^{\left(1\right)}\left(x\right)=j_l\left(x\right)+i{n}_l\left(x\right)$

Thus,

$$\begin{gathered} e^{i{\mathrm{δ}}_l}\sin {\mathrm{δ}}_l=i\frac{j_l\left(ka\right)}{j_l\left(x\right)+i{n}_l\left(x\right)} \\ {e}^{i{\mathrm{δ}}_l}\sin {\mathrm{δ}}_l=i\frac{1}{1+i\left(n/j\right)} \\ {e}^{i{\mathrm{δ}}_l}\sin {\mathrm{δ}}_l=i\frac{1-i\left(n/j\right)}{1+i{\left(n/j\right)}^2} \\ {e}^{i{\mathrm{δ}}_l}\sin {\mathrm{δ}}_l=\frac{\left(n/j\right)+i}{1+i{\left(n/j\right)}^2} \end{gathered}$$

But,

Thus,

$\cos \left({\mathrm{δ}}_l\right)\sin \left({\mathrm{δ}}_l\right)+i{\sin }^2\left({\mathrm{δ}}_l\right)=\frac{\left(n/j\right)+i}{1+{\left(n/j\right)}^2}$

By equating the real and imaginary parts we get:

$$\begin{gathered} \cos \left({\mathrm{δ}}_l\right)\sin \left({\mathrm{δ}}_l\right)=\frac{\left(n/j\right)}{1+{\left(n/j\right)}^2} \\ {\sin }^2\left({\mathrm{δ}}_l\right)=\frac{1}{1+{\left(n/j\right)}^2} \end{gathered}$$

Dividing the second equation by the first one, so we get:

$$\begin{gathered} {\mathrm{δ}}_l=\frac{j}{n} \\ {\mathrm{δ}}_l={\tan }^{-1}\left(\frac{j_l\left(ka\right)}{n_l\left(ka\right)}\right) \end{gathered}$$