Question

An \(\alpha\) -particle of energy \(5 \mathrm{MeV}\) is scattered though \(180^{\circ}\) by a fixed uranium nucleus. The distance of the closet approach is of the order of (A) \(10^{-8} \mathrm{~cm}\) (B) \(10^{-12} \mathrm{~cm}\) (C) \(10^{-10} \mathrm{~cm}\) (D) \(10^{-15} \mathrm{~cm}\)

Short answer

The distance of closest approach is of the order of \(10^{-12} cm\). So the answer is (B).

Step-by-step solution

Identify the given information

We are given the following information: - The energy of the alpha particle, E = 5 MeV. We need to convert this to Joules: \(E = 5 \times 10^6 eV \times 1.6 \times 10^{-19} J/eV = 8 \times 10^{-13} J\). - The scattering angle is 180 degrees. - The charge of the alpha particle is 2e, where e is the electron charge, and the charge of the uranium nucleus is 92e (since uranium has an atomic number of 92).

Calculate the electrical potential energy

First, we need to find the electrical potential energy, U, between the alpha particle and the uranium nucleus at the distance of closest approach, d. The formula for electrical potential energy is: \(U = \dfrac{kq_1q_2}{d}\) Where: - k is the Coulomb constant, \(k = 8.99 \times 10^9 Nm^2/C^2\) - \(q_1\) is the charge of the alpha particle, \(q_1 = 2e = 3.2 \times 10^{-19} C\) - \(q_2\) is the charge of the uranium nucleus, \(q_2 = 92e = 1.472 \times 10^{-17} C\)

Use conservation of energy

The law of conservation of energy states that the total energy before and after must stay constant. Since the initial energy was purely kinetic, and at the distance of closest approach all kinetic energy is converted to potential energy, we have: Initial kinetic energy = Final potential energy \(E = U\)

Solve for the distance of closest approach

Now that we have the equation E = U (with E and U having been already calculated), we can solve for d. Substitute \(E\) and \(U\) with the energy and electrical potential energy: \(8 \times 10^{-13} J = \dfrac{(8.99 \times 10^9 Nm^2/C^2)(3.2 \times 10^{-19} C)(1.472 \times 10^{-17} C)}{d}\) Now, isolate "d": \(d = \dfrac{(8.99 \times 10^9 Nm^2/C^2)(3.2 \times 10^{-19} C)(1.472 \times 10^{-17} C)}{8 \times 10^{-13} J}\) And finally, calculate the value for the distance of closest approach: \(d \approx 1.7 \times 10^{-14} m\) Converting this to centimeters: \(d \approx 1.7 \times 10^{-12} cm\) Checking the available options, the distance of the closest approach is of the order of \(10^{-12} cm\). So the answer is (B).