Question

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy \(\mathrm{K}_{1}\) is ro. The distance of the closest approach when the \(\alpha\) - particle is fired at the same nucleus with kinetic energy \(2 \mathrm{k}_{1}\) will be. (A) \(\left(\mathrm{r}_{0} / 2\right)\) (B) \(4 r_{0}\) (C) \(\left(\mathrm{r}_{0} / 4\right)\) (D) \(2 \mathrm{r}_{0}\)

Short answer

The distance of the closest approach when the alpha particle is fired at the same nucleus with kinetic energy \(2K_1\) will be \(r' = 2r_0\).

Step-by-step solution

Write down the known variables and equations

Let's make a list of the known variables: - Initial kinetic energy is \(K_{1}\). - Distance of closest approach with initial kinetic energy \(K_1\) is \(r_0\). - The charge of an alpha particle is \(q_{\alpha} = 2e\), where \(e\) is the elementary charge. - The charge of the nucleus is \(q_n\), which will be multiplied with the elementary charge constant. - The electrostatic force between the alpha particle and the nucleus is given by Coulomb's law: \(F = \frac{kq_{\alpha}q_n}{r^2}\), where k is the electrostatic constant and r is the distance between the charges. Now, let's write down the energy conservation equation: \(K_1 + U_1 = K_2 + U_2\), where \(K_1\) and \(K_2\) are the initial and final kinetic energies, and \(U_1\) and \(U_2\) are the initial and final potential energies. At the closest approach, the kinetic energy becomes zero (\(K_2 = 0\)) and we have: \(K_1 + \frac{kq_{\alpha}q_n}{r_1} = \frac{kq_{\alpha}q_n}{r_2}\), When the initial kinetic energy is doubled (2K1), the equation becomes: \(2K_1 + \frac{kq_{\alpha}q_n}{r'_1} = \frac{kq_{\alpha}q_n}{r'_2}\), Where \(r'_1\) and \(r'_2\) are the new initial and final distances of closest approach.

Write down the equations for the distances of closest approach

Using the energy conservation equation with the initial and doubled kinetic energies, we can rewrite the equations for the distances of closest approach: For the initial kinetic energy \(K_1\): \(\frac{kq_{\alpha}q_n}{r_0} = K_1\), For the doubled kinetic energy \(2K_1\): \(\frac{kq_{\alpha}q_n}{r'} = 2K_1\).

Combine the equations and solve for the new distance of closest approach

Dividing the equations (\(\frac{kq_{\alpha}q_n}{r'} = 2K_1\) and \(\frac{kq_{\alpha}q_n}{r_0} = K_1\)), we get: \(\frac{r_0}{r'} = \frac{1}{2}\), Hence, \(r' = 2r_0\). So, the distance of the closest approach when the alpha particle is fired at the same nucleus with kinetic energy 2\(K_1\) will be: \(r' = 2r_0\). The correct answer is option (D).