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We need to calculate total cross-section for scattering using first Born approximation for given potential

$\mathrm{V}(\mathrm{r}鈨�)={\mathrm{Ae}}^{-{\mathrm{渭谤}}^2}$

Potential has spherical symmetry, so in order to find scattering amplitude we use following expression:

$\mathrm{f}\left(\mathrm{蠎}\right)=\frac{-2\mathrm{m}}{{鈩�}^2\mathrm{魏}}{鈭�}_0^{鈭�}\mathrm{r}\mathrm{V}\left(\mathrm{r}\right)\sin \left(\mathrm{魏谤}\right)\mathrm{dr}$ 鈥︹��.. (1)

$\mathrm{f}\left(\mathrm{蠎}\right)=\frac{-2\mathrm{m}}{{鈩�}^2\mathrm{魏}}{鈭�}_0^{鈭�}\mathrm{r}{\mathrm{Ae}}^{-{\mathrm{渭谤}}^2}\sin \left(\mathrm{魏谤}\right)\mathrm{dr}$ 鈥︹��.. (2)

To take the constant outside,

$\mathrm{f}\left(\mathrm{蠎}\right)=\frac{-2\mathrm{mA}}{{鈩�}^2\mathrm{魏}}{鈭�}_0^{鈭�}\mathrm{r}{\mathrm{e}}^{-{\mathrm{渭谤}}^2}\sin \left(\mathrm{魏谤}\right)\mathrm{dr}鈥�鈥�..\left(3\right)$ $\mathrm{f}\left(\mathrm{蠎}\right)=\frac{-2\mathrm{mA}}{{鈩�}^2\mathrm{魏}}{鈭�}_0^{鈭�}\frac{\mathrm{d}}{\mathrm{dr}}\frac{-1}{2\mathrm{渭}}{\mathrm{e}}^{-{\mathrm{渭谤}}^2}\sin \left(\mathrm{魏谤}\right)\mathrm{dr}......\left(4\right)$

To integrate,

$\mathrm{f}\left(\mathrm{蠎}\right)=\frac{\mathrm{mA}}{\mathrm{渭}{鈩�}^2\mathrm{魏}}[{\mathrm{e}}^{-{\mathrm{渭谤}}^2}\sin \mathrm{魏谤}{|}_0^{鈭�}-\mathrm{魏}{鈭�}_0^{鈭�}{\mathrm{e}}^{-{\mathrm{渭谤}}^2}\cos \mathrm{魏}\mathrm{r}\mathrm{dr}]........\left(5\right)$

$\mathrm{f}\left(\mathrm{蠎}\right)=\frac{-\mathrm{mA}}{\mathrm{渭}{鈩�}^2}\frac{\sqrt{\mathrm{蟺}}}{2\sqrt{\mathrm{渭}}}{\mathrm{e}}^{-{\mathrm{魏}}^2/4\mathrm{渭}}.......\left(6\right)$

$f\left(蠎\right)=\frac{-mA\sqrt{蟺}}{2{渭}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈩�}^2}{e}^{-{魏}^2/4渭}......\left(7\right)$

Where in equation (5) we used integration by parts. Only has angular dependence:

$\mathrm{魏}=2\mathrm{k}\sin \frac{\mathrm{蠎}}{2}$

We can now calculate the total cross-section:

$$\begin{gathered} \frac{d蟽}{d惟}=\left|f\right(蠎){|}^2 \\ \frac{d蟽}{d惟}=\frac{蟺m^2{A}^2}{4{鈩�}^4{渭}^3}{e}^{-{魏}^2/2渭} \end{gathered}$$

Now calculate the total cross-section,

role="math" localid="1658388382186" $$\begin{gathered} \mathrm{蟽}=(\frac{{\mathrm{蟺尘}}^2{\mathrm{A}}^2}{4{鈩�}^4{\mathrm{渭}}^3}{鈭�}^{0\mathrm{蟺}鈭�{鈭�}_0^{2\mathrm{蟺}}\mathrm{d}-4{\mathrm{k}}^2{\sin }^2\mathrm{蠎}/2/2\mathrm{渭}}\sin \\ \mathrm{蟽}=\frac{{\mathrm{蟺}}^2{\mathrm{m}}^2{\mathrm{A}}^2}{2{鈩�}^4{\mathrm{渭}}^3}{鈭�}^{0\mathrm{蟺}鈭�-2{\mathrm{k}}^2{\sin }^2\frac{\mathrm{蠎}}{2}/\mathrm{渭}}\sin \\ \mathrm{蟽}=\frac{{\mathrm{蟺}}^2{\mathrm{m}}^2{\mathrm{A}}^2}{2{鈩�}^4{\mathrm{渭}}^3}{鈭�}_0^{\mathrm{蟺}}2\cos \frac{\mathrm{蠎}}{2}\sin \frac{\mathrm{蠎}}{2}\mathrm{e}-2{\mathrm{k}}^2{\sin }^2\frac{\mathrm{蠎}}{2}/\mathrm{渭诲蠎} \end{gathered}$$

To integrate,

$$\begin{gathered} \mathrm{u}=\frac{2{\mathrm{k}}^2}{\mathrm{渭}}\sin \frac{\mathrm{蠎}}{2}鈥�鈥�鈥�0鈫�0鈥�鈥�鈥�\mathrm{蟺}鈫�\frac{2{\mathrm{k}}^2}{\mathrm{渭}} \\ du=\frac{2k^2}{渭}sin\frac{蠎}{2}cos\frac{蠎}{2}d蠎 \\ 蟽=(\frac{{蟺}^2{m}^2{A}^2}{{鈩�}^4{渭}^3}{鈭�}_0^{2k^2/渭}{e}^{-渭}du \\ 蟽=\frac{{蟺}^2{m}^2{A}^2}{2k^2{鈩�}^4{渭}^3}[1-e^{-2k^2/渭}]. \end{gathered}$$