(a) Calculate the differential cross section in the Born approximation for the potential V(r)=V_{0} e^{-r / R} / r, known as the Yukawa potential.
(b) Calculate the total cross section. (c) Find the relation between V_{0} and R so that the Born approximation is valid.
(a) Inserting V(r)=V_{0} e^{-r / R} / r into (11.70),
\frac{d \sigma}{d \Omega}=|f(\theta)|^{2}=\frac{4 \mu^{2}}{\hbar^{4} q^{2}}\left|\int_{0}^{\infty} r^{\prime} V\left(r^{\prime}\right) \sin \left(q r^{\prime}\right) d r^{\prime}\right|^{2}.       (11.70)
we obtain
\frac{d \sigma}{d \Omega}=\frac{4 \mu^{2} V_{0}^{2}}{\hbar^{4} q^{2}}\left|\int_{0}^{\infty} e^{-r / R} \sin (q r) d r\right|^{2},         (11.133)
where
\int_{0}^{\infty} e^{-r / R} \sin (q r) d r=\frac{1}{2 i} \int_{0}^{\infty} e^{-(1 / R-i q) r} d r-\frac{1}{2 i} \int_{0}^{\infty} e^{-(1 / R+i q) r} d r
=\frac{1}{2 i}\left[\frac{1}{1 / R-i q}-\frac{1}{1 / R+i q}\right]=\frac{q}{1 / R^{2}+q^{2}};Â Â Â Â Â Â Â Â Â Â Â Â (11.134)
hence
\frac{d \sigma}{d \Omega}=\frac{4 \mu^{2} V_{0}^{2}}{\hbar^{4}} \frac{1}{\left(1 / R^{2}+q\right)^{2}}=\frac{4 \mu^{2} V_{0}^{2}}{\hbar^{4}} \frac{1}{\left[1 / R^{2}+4 k^{2} \sin ^{2}(\theta / 2)\right]^{2}}.        (11.135)
Note that a connection can be established between this relation and the differential cross section for a Coulomb potential V(r)=Z_{1} Z_{2} e^{2} / r. For this, we need only to insert V_{0}=-Z_{1} Z_{2} e^{2} into (11.135) and then take the limit R → ∞; this leads to (11.77):
\frac{d \sigma}{d \Omega}=\left(\frac{2 Z_{1} \mu Z_{2} e^{2}}{\hbar^{2} q^{2}}\right)^{2}=\left(\frac{Z_{1} Z_{2} \mu e^{2}}{2 \hbar^{2} k^{2}}\right)^{2} \sin ^{-4}\left(\frac{\theta}{2}\right)=\frac{Z_{1}^{2} Z_{2}^{2} e^{4}}{16 E^{2}} \sin ^{-4}\left(\frac{\theta}{2}\right),       (11.77)
\left(\frac{d \sigma}{d \Omega}\right)_{R u t h e r f o r d}=\lim _{R \rightarrow \infty}\left(\frac{d \sigma}{d \Omega}\right)_{\text {Yukawa }} .          (11.136)
(b) The total cross section can be obtained at once from (11.135):
\sigma=\int \frac{d \sigma}{d \Omega} \sin \theta d \theta d \varphi=2 \pi \int_{0}^{\pi} \frac{d \sigma}{d \Omega} \sin \theta d \theta=2 \pi \frac{4 \mu^{2} V_{0}^{2} R^{4}}{\hbar^{4}} \int_{0}^{\pi} \frac{\sin \theta d \theta}{\left(1+4 k^{2} R^{2} \sin ^{2}(\theta / 2)\right)^{2}} .     (11.137)
The change of variable x=2 k R \sin (\theta / 2) leads to \sin \theta d \theta=x d x /\left(k^{2} R^{2}\right); hence
\sigma=\frac{8 \pi \mu^{2} V_{0}^{2} R^{4}}{\hbar^{4}} \frac{1}{k^{2} R^{2}} \int_{0}^{2 k R} \frac{x d x}{\left(1+x^{2}\right)^{2}}=\frac{16 \pi \mu^{2} V_{0}^{2} R^{4}}{\hbar^{4}} \frac{1}{1+4 k^{2} R^{2}}
=\frac{16 \pi \mu^{2} V_{0}^{2} R^{4}}{\hbar^{4}} \frac{1}{1+8 \mu E R^{2} / \hbar^{2}},          (11.138)
where we have used k^{2}=2 \mu E / \hbar^{2} ; E is the energy of the scattered particle.
(c) The validity condition of the Born approximation is
\frac{\mu V_{0}}{\hbar^{2} k^{2}}\left|\int_{0}^{\infty} \frac{e^{-a r}}{r}\left(e^{2 i k r}-1\right) d r\right| \ll 1,       (11.139)
where a = 1/R. To evaluate the integral
I=\int_{0}^{\infty} \frac{e^{-a r}}{r}\left(e^{2 i k r}-1\right) d r         (11.140)
let us differentiate it with respect to the parameter a:
\frac{\partial I}{\partial a}=-\int_{0}^{\infty} e^{-a r}\left(e^{2 i k r}-1\right) d r=-\frac{1}{a-2 i k}+\frac{1}{a}.       (11.141)
Now, integrating over the parameter a such that I (a = +∞) = 0, we obtain
I=\ln a-\ln (a-2 i k)=-\ln \left(1-2 i \frac{k}{a}\right)=-\frac{1}{2} \ln \left(1+\frac{4 k^{2}}{a^{2}}\right)+i \tan ^{-1}\left(\frac{2 k}{a}\right) .        (11.142)
Thus, the validity condition (11.139) becomes
\frac{\mu V_{0}}{\hbar^{2} k^{2}}\left\{\frac{1}{4}\left[\ln \left(1+4 k^{2} R^{2}\right)\right]^{2}+\left(\tan ^{-1}(2 k R)\right)^{2}\right\}^{1 / 2} \ll 1.          (11.143)