An \(\alpha\) particle with a kinetic energy of \(1.0 \mathrm{MeV}\) is headed straight toward a gold nucleus. (a) Find the distance of closest approach between the centers of the \(\alpha\) particle and gold nucleus. (Assume the gold nucleus remains stationary. since its mass is much larger than that of the \(\alpha\) particle, this assumption is a fairly good approximation.) (b) Will the two get close enough to "touch"? (c) What is the minimum initial kinetic energy of an \(\alpha\) particle that will make contact with the gold nucleus?

We will first find the electrostatic force acting on the α particle due to the gold nucleus. The equation governing electrostatic force is Coulomb's law, given by: $$F = k\frac{q_1q_2}{r^2}$$ Where: - \(F\) is the electrostatic force, - \(k\) is the electrostatic constant (\(8.9875 \times 10^9 \mathrm{Nm^2C^{-2}}\)), - \(q_1\) and \(q_2\) are the charges of the two interacting particles, - \(r\) is the distance between the centers of the particles. The α particle has a charge of \(2e\), and the gold nucleus has a charge of \(79e\). Thus, the electrostatic force is: $$F = k\frac{(2e)(79e)}{r^2}$$

At the distance of closest approach, the kinetic energy of the α particle becomes zero, and all energy is converted into potential energy due to the electrostatic force. Therefore, we can use the conservation of energy principle to find the closest approach distance: $$K_{initial} = U_{final}$$ Where \(K_{initial}\) is the initial kinetic energy of the α particle, and \(U_{final}\) is the final electrostatic potential energy. Using the equation of electrostatic potential energy \(U = k \frac{q_1q_2}{r}\), we can write the potential energy at the final position as: $$U_{final} = k\frac{(2e)(79e)}{r}$$ Now, we can substitute the given values into the conservation of energy equation: $$1.0 \times 10^6 eV = k\frac{(2e)(79e)}{r}$$ and solve for \(r\): $$r = k\frac{(2e)(79e)}{1.0 \times 10^6 eV}$$ Converting the result to meters, we find: $$r \approx 4.6 \times 10^{-14} m$$

To determine if the α particle and the gold nucleus will "touch," we compare the closest approach distance calculated in Step 2 with the sum of their respective radii. The radius of the α particle is approximately \(1.2 \times 10^{-15} m\), and the radius of a gold nucleus is approximately \(7.3 \times 10^{-15} m\). The sum of their radii is: $$r_{\alpha} + r_{Au} = 1.2 \times 10^{-15} m + 7.3 \times 10^{-15} m = 8.5 \times 10^{-15} m$$ Since the distance of closest approach (\(4.6 \times 10^{-14} m\)) is significantly larger than the sum of their radii (\(8.5 \times 10^{-15} m\)), the α particle and gold nucleus will not "touch."

To find the minimum initial kinetic energy required for contact, we use the conservation of energy principle again. However, this time, we set the closest approach distance equal to the sum of their radii: $$K_{initial} = k\frac{(2e)(79e)}{r_{\alpha} + r_{Au}} = k\frac{(2e)(79e)}{8.5 \times 10^{-15} m}$$ Converting the result to electron volts, we find: $$K_{initial} \approx 1.9 \times 10^7 eV$$ Therefore, the minimum initial kinetic energy required for an α particle to make contact with the gold nucleus is approximately \(1.9 \times 10^7 eV\).