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A narrow beam of protons with kinetic energy \(T=1.4 \mathrm{MeV}\) falls normally on a brass foil whose mass thickness \(\rho d=1.5 \mathrm{mg} / \mathrm{cm}^{2}\). The weight ratio of copper and zinc in the foil is equal to 7: 3 respectively. Find the fraction of the protons scattered through the angles exceeding \(\theta_{0}=30^{\circ} .\)
\[ \text { 6.15. } \Delta N / N=\frac{\pi e^{4}}{4 T^{2}}\left(0.7 \frac{Z_{1}^{2}}{M_{1}}+0.3 \frac{Z_{2}^{2}}{M_{2}}\right) \rho d N_{A} \cot ^{2} \frac{\theta}{2}=1.4 \cdot 10^{-3} \] where \(Z_{1}\) and \(Z_{2}\) are the atomic numbers of copper and zinc, \(M_{1}\) and \(M_{2}\) are their molar masses, \(N_{A}\) is Avogadro's number.