Determine the classical differential cross section for Rutherford scattering of alpha particles of energy \(5.00 \mathrm{MeV}\) projected at uranium atoms at an angle of \(35.0^{\circ}\) from the initial direction. Assume point charges for both the target and the projectile atoms.

We are given the energy of the alpha particles in MeV. To convert this to Joules, we can use the conversion factor we mentioned earlier. So we have: $$ E = 5.00\,\mathrm{MeV} \cdot \frac{1\,\mathrm{u}}{931.5\,\mathrm{MeV}} \cdot \frac{1.66054\times10^{-27}}{1.60219\times10^{-13}}\,\mathrm{J} \approx 1.6022\times10^{-13}\,\mathrm{J} $$

The wave number, k, can be calculated using the relation \(E = \frac{d^2 |\mathbf{p}|}{2 m_\alpha}=\frac{d^2 \hbar^2 k^2}{2dm_\ alpha}=\hbar^2k^2/2m_\alpha\), where \( d=1\): $$ k^2 = \frac{2m_\alpha E}{\hbar^2} $$ where \(m_\alpha = 4\,\mathrm{u} \cdot 1.66054\times10^{-27}\,\mathrm{kg/u} = 6.64216\times10^{-27}\,\mathrm{kg}\). The reduced Planck's constant, \(\hbar\), is equal to \(1.055\times10^{-34}\,\mathrm{Js}\). So we have: $$ k^2 = \frac{2\cdot 6.64216\times10^{-27}\,\mathrm{kg} \cdot 1.6022\times10^{-13}\,\mathrm{J}}{(1.055\times10^{-34}\,\mathrm{Js})^2} \approx 1.129\times10^{25}\,\mathrm{m^{-2}} $$

Now, we can use the Rutherford scattering formula to calculate the classical differential cross section: $$ \frac{d\sigma}{d\Omega} = \left(\frac{Ze^2}{8\pi\epsilon_0 k^2}\right)^2 \frac{1}{\sin^4(\theta/2)} $$ where \(Z=Z_\alpha \cdot Z_U = 2 \cdot 92 = 184\), \(e = 1.602\times10^{-19}\,\mathrm{C}\), \(\theta = 35.0^\circ\), and \(\epsilon_0 = 8.85\times10^{-12}\,\mathrm{C^2/Nm^2}\). We first compute the sine of half of the scattering angle: $$ \sin^4(\theta/2) = \sin^4(35.0^\circ/2) \approx 0.0255 $$ Now, we calculate the classical differential cross section: $$ \frac{d\sigma}{d\Omega} = \left(\frac{184 \cdot (1.602\times10^{-19}\,\mathrm{C})^2}{8\pi(8.85\times10^{-12}\,\mathrm{C^2/Nm^2})(1.129\times10^{25}\,\mathrm{m^{-2}})}\right)^2 \frac{1}{0.0255} \approx 6.397\times10^{-30}\,\mathrm{m^2/sr} $$ The classical differential cross section for Rutherford scattering of alpha particles of energy \(5.00 \,\mathrm{MeV}\) projected at uranium atoms at an angle of \(35.0^{\circ}\) from the initial direction is approximately \(6.397\times10^{-30}\,\mathrm{m^2/sr}\).