91Ó°ÊÓ

The Yukawa potential in its general form is:

$V\left(r\right)=\mathrm{β}\frac{e^{-\mathrm{Ó\neg }}}{r}$

Where $\mathrm{β}$and are $\mathrm{μ}$constants.

The Born approximation gives:

$f\left(\mathrm{θ}\right)=-\frac{2m\mathrm{β}}{\sout{h^2}{\left({\mathrm{μ}}^2+k^2\right)}^2}$

The differential cross section is the square modulus of the amplitude scattering, that is:

role="math" localid="1658378980970" $$\begin{gathered} D\left(\mathrm{θ}\right)={\left|f\left(\mathrm{θ}\right)\right|}^2 \\ D\left(\mathrm{θ}\right)={\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{(u^2+k^2)}^2} \end{gathered}$$

Where;

$k=2k\sin \frac{\mathrm{θ}}{2}$

Thus,

$D\left(\mathrm{θ}\right)={\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{\left(u^2+{\left(2k\right)}^2{\sin }^2{\left(\mathrm{θ}/2\right)}^2\right)}$ .........(1)

The total cross section is given by:

$\mathrm{δ}=∫D\left(\mathrm{θ}\right)\sin \left(\mathrm{θ}\right)d\mathrm{θ}d\mathrm{ϕ}$

Substitute from (1) we get:

$\mathrm{δ}=\frac{4m^2{\mathrm{β}}^2}{\sout{h^2}}{∫}_0^{\mathrm{π}}d\mathrm{θ}{∫}_0^{2\mathrm{π}}d\mathrm{ϕ}\frac{\sin \mathrm{θ}}{{\left(k^{2+}{\left(2k\right)}^2{\sin }^2\left(\mathrm{θ}/2\right)\right)}^2}$

First integrate forthen we use the identity,

$\sin \left(\mathrm{θ}\right)=2\sin \left(\frac{\mathrm{θ}}{2}\right)\cos \left(\frac{\mathrm{θ}}{2}\right)$

So, we get:

$\mathrm{δ}=4\mathrm{π}{\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\mathrm{μ}}^4}{∫}_0^{\mathrm{π}}\frac{1}{{\left[1+{\left(2k/\mathrm{μ}\right)}^2{\sin }^2\left(\mathrm{θ}/2\right)\right]}^2}\sin \left(\frac{\mathrm{θ}}{2}\right)\cos \left(\frac{\mathrm{θ}}{2}\right)d\mathrm{θ}$

Let localid="1658819683791" $x=\frac{2k}{\mathrm{μ}}\sin \left(\mathrm{θ}/2\right),$then $\frac{\mathrm{μ}}{k}x=2\sin \left(\mathrm{θ}/2\right)and\frac{\mathrm{μ}}{k}dx=\cos \left(\mathrm{θ}/2\right)d\mathrm{θ}$, so we get:

$\mathrm{δ}=2\mathrm{π}{\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\mathrm{μ}}^4}{\left(\frac{\mathrm{μ}}{k}\right)}^2∫\frac{x}{{\left(1+x^2\right)}^2}dx$

Note that we must change the limits of the integral to be:

$$\begin{gathered} \mathrm{θ}=0 \\ x=0 \\ \mathrm{θ}=\mathrm{π} \\ x=2k/\mathrm{μ} \end{gathered}$$

Thus:

$\mathrm{σ}=2\mathrm{π}{\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\mathrm{μ}}^4}{\left(\frac{\mathrm{μ}}{k}\right)}^2{∫}_0^{2k/\mathrm{μ}}\frac{x}{\left(1+x^2\right)}dx$

The integral now is easy to solve. We get:

role="math" localid="1658382645180" $$\begin{gathered} \mathrm{σ}=2\mathrm{π}{\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\left(\mathrm{μ}k\right)}^2}\left[-\frac{1}{2}\frac{1}{\left(1+x^2\right)}\right]\overline{){}_0^{2k/\mathrm{μ}}} \\ \mathrm{σ}=\mathrm{π}{\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\left(\mathrm{μ}k\right)}^2}\left[1-\frac{1}{1+{\left(2k/\mathrm{μ}\right)}^2}\right] \\ \mathrm{σ}=\mathrm{π}{\left(\frac{2m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\left(\mathrm{μ}k\right)}^2}\left[\frac{4{\left(k/\mathrm{μ}\right)}^2}{1+4k^2/{\mathrm{μ}}^2}\right] \\ \mathrm{σ}=\mathrm{π}{\left(\frac{4m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\left(\mathrm{μ}k\right)}^2}\frac{1}{{\mathrm{μ}}^2+4k^2} \end{gathered}$$

But,

$k^2=\frac{2mE}{\sout{h^2}}$

Hence,

$\mathrm{σ}=\mathrm{π}{\left(\frac{4m\mathrm{β}}{\sout{h^2}}\right)}^2\frac{1}{{\left(\mathrm{μ}k\right)}^2+8mE}$.