Calculate the total cross-section for scattering from a Yukawa potential, in the Born approximation. Express your answer as a function of E.

The Yukawa potential in its general form is:

$V\left(r\right)=\beta \frac{e^{-\omega }}{r}$

Where $\beta$and are $\mu$constants.

The Born approximation gives:

$f\left(\theta \right)=-\frac{2m\beta }{\sout{h^2}{\left({\mu }^2+k^2\right)}^2}$

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

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

Where;

$k=2k\sin \frac{\theta }{2}$

Thus,

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

The total cross section is given by:

$\delta =\int D\left(\theta \right)\sin \left(\theta \right)d\theta d\varphi$

Substitute from (1) we get:

$\delta =\frac{4m^2{\beta }^2}{\sout{h^2}}{\int }_0^{\pi }d\theta {\int }_0^{2\pi }d\varphi \frac{\sin \theta }{{\left(k^{2+}{\left(2k\right)}^2{\sin }^2\left(\theta /2\right)\right)}^2}$

First integrate forthen we use the identity,

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

So, we get:

$\delta =4\pi {\left(\frac{2m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\mu }^4}{\int }_0^{\pi }\frac{1}{{\left[1+{\left(2k/\mu \right)}^2{\sin }^2\left(\theta /2\right)\right]}^2}\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)d\theta$

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

$\delta =2\pi {\left(\frac{2m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\mu }^4}{\left(\frac{\mu }{k}\right)}^2\int \frac{x}{{\left(1+x^2\right)}^2}dx$

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

$$\begin{gathered} \theta =0 \\ x=0 \\ \theta =\pi \\ x=2k/\mu \end{gathered}$$

Thus:

$\sigma =2\pi {\left(\frac{2m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\mu }^4}{\left(\frac{\mu }{k}\right)}^2{\int }_0^{2k/\mu }\frac{x}{\left(1+x^2\right)}dx$

The integral now is easy to solve. We get:

role="math" localid="1658382645180" $$\begin{gathered} \sigma =2\pi {\left(\frac{2m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\left(\mu k\right)}^2}\left[-\frac{1}{2}\frac{1}{\left(1+x^2\right)}\right]\overline{){}_0^{2k/\mu }} \\ \sigma =\pi {\left(\frac{2m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\left(\mu k\right)}^2}\left[1-\frac{1}{1+{\left(2k/\mu \right)}^2}\right] \\ \sigma =\pi {\left(\frac{2m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\left(\mu k\right)}^2}\left[\frac{4{\left(k/\mu \right)}^2}{1+4k^2/{\mu }^2}\right] \\ \sigma =\pi {\left(\frac{4m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\left(\mu k\right)}^2}\frac{1}{{\mu }^2+4k^2} \end{gathered}$$

But,

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

Hence,

$\sigma =\pi {\left(\frac{4m\beta }{\sout{h^2}}\right)}^2\frac{1}{{\left(\mu k\right)}^2+8mE}$.