Question

An alpha particle (mass \(=4.00 \mathrm{u}\) ) has a head-on, elastic collision with a nucleus (mass \(=166 \mathrm{u}\) ) that is initially at rest. What percentage of the kinetic energy of the alpha particle is transferred to the nucleus in the collision?

Short answer

Answer: To calculate the percentage of kinetic energy transferred, we need the initial or final velocities of the alpha particle, which were not provided in this exercise. If such values were provided or could be calculated from given information, the remaining calculations could be completed using the conservation of momentum and conservation of energy equations to find the percentage of kinetic energy transferred.

Step-by-step solution

Conservation of Momentum

According to the law of conservation of momentum, the total momentum of the system before the collision equals the total momentum of the system after the collision. Let's denote the masses of alpha particle and nucleus as m1 and m2 respectively, and their initial and final velocities as u1, u2, v1, and v2 respectively. Since the nucleus is initially at rest, u2 = 0. Initially, Momentum of alpha particle = m1u1 Initially, Momentum of nucleus = m2u2 (which is 0) Finally, Momentum of alpha particle = m1v1 Finally, Momentum of nucleus = m2v2 Conservation of momentum equation can be written as: m1u1 + m2u2 = m1v1 + m2v2 Plugging in the known values, we get: m1u1 = m1v1 + m2v2

Conservation of Energy

For an elastic collision, the total mechanical energy is conserved. Hence, the initial kinetic energy (KE_initial) equals the final kinetic energy (KE_final). KE_initial = KE_final Initial kinetic energy of alpha particle = \(\frac{1}{2}m1u1^2\) Initial kinetic energy of nucleus = 0 (since it's initially at rest) Final kinetic energy of alpha particle = \(\frac{1}{2}m1v1^2\) Final kinetic energy of nucleus = \(\frac{1}{2}m2v2^2\) Now, equating the initial and final kinetic energies, we get: \(\frac{1}{2}m1u1^2 = \frac{1}{2}m1v1^2 + \frac{1}{2}m2v2^2\)

Solve for v2

To find the percentage of kinetic energy transferred, we need to find the final velocity of the nucleus (v2) after the collision. Using the conservation of momentum equation from Step 1, we can solve for v2: v2 = \(\frac{m1(u1-v1)}{m2}\)

Substitute values and find the percentage of transferred kinetic energy

The mass of alpha particle m1 = 4.00 u and mass of nucleus m2 = 166 u. Calculate the final kinetic energy of the nucleus by substituting the values: KE_final_nucleus = \(\frac{1}{2}m2v2^2\) Percentage of kinetic energy transferred = \(\frac{KE_{final\_nucleus}}{KE_{initial}}\) × 100 Plug in the values and compute the final answer. Note that we have not been provided the initial or final velocities of the alpha particle. If such values were provided or could be calculated from given information, the remaining calculations could be completed to find the percentage of kinetic energy transferred.