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

Step 1: Write the conservation of momentum and conservation of kinetic energy equations: \(m_{\alpha}v_{\alpha} = m_{\alpha}v_{\alpha}' + m_Nv_N'\) (momentum) \(\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \frac{1}{2}m_{\alpha}v_{\alpha}'^2 + \frac{1}{2}m_Nv_N'^2\) (kinetic energy) Step 2: Solve the equations simultaneously to find the final velocities of the alpha particle, \(v_{\alpha}'\), and the nucleus, \(v_N'\). Step 3: Calculate the final kinetic energies and compare them to the initial kinetic energies to find the percentage of energy transferred: Percentage of energy transferred = \(\frac{E_{\alpha} - E_{\alpha}'}{E_{\alpha}} \times 100\) Answer: The percentage of kinetic energy transferred from the alpha particle to the nucleus in this specific elastic head-on collision is ________%.

Step-by-step solution

Identify the information given and the formula required for the solution.

The information given to us is: - The mass of the alpha particle (\(m_{\alpha}\)) is \(4.00\mathrm{u}\). - The mass of the nucleus (\(m_N\)) is \(166\mathrm{u}\). - The collision is elastic and head-on. In an elastic collision, the principle of conservation of momentum is followed. This means that the total momentum before the collision is equal to the total momentum after the collision. Since the nucleus is initially at rest, its momentum is 0. Therefore, the total momentum before the collision is equal to the momentum of the alpha particle. Mathematically, the conservation of momentum can be written as: \(m_{\alpha}v_{\alpha} = m_{\alpha}v_{\alpha}' + m_Nv_N'\), where: - \(v_{\alpha}\) is the initial velocity of the alpha particle - \(v_{\alpha}'\) is the final velocity of the alpha particle - \(v_N'\) is the final velocity of the nucleus Additionally, in an elastic collision, the principle of conservation of kinetic energy is followed. So, we can write the conservation of kinetic energy as: \(\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \frac{1}{2}m_{\alpha}v_{\alpha}'^2 + \frac{1}{2}m_Nv_N'^2\) Now, we need to solve the momentum and kinetic energy conservation equations to find the final velocities.

Solve the momentum and kinetic energy equations to find the final velocities.

First, let's divide both sides of the momentum equation by \(m_{\alpha}\), which gives us: \(v_{\alpha} =v_{\alpha}' + \frac{m_N}{m_{\alpha}}v_N'\) Now, we have one equation with two unknowns, \(v_{\alpha}'\) and \(v_N'\). We need to use the kinetic energy equation to find another equation. Divide both sides of the kinetic energy equation by \(\frac{1}{2}m_{\alpha}\), which gives us: \(v_{\alpha}^2 =v_{\alpha}'^2 + \frac{m_N}{m_{\alpha}}v_N'^2\) Since we have two equations and two unknowns, we can solve them simultaneously. Let's substitute the value of \(v_{\alpha}'\) from the momentum equation to the kinetic energy equation: \(v_{\alpha}^2 = (v_{\alpha}-\frac{m_N}{m_{\alpha}}v_N')^2 + \frac{m_N}{m_{\alpha}}v_N'^2\) After solving this equation, we find the final velocities of the alpha particle, \(v_{\alpha}'\), and the nucleus, \(v_N'\).

3: Find the final kinetic energies and compare them to the initial kinetic energies.

Once we have the final velocities, we can find the final kinetic energies of the alpha particle and the nucleus: - Final kinetic energy of the alpha particle: \(E_{\alpha}' = \frac{1}{2}m_{\alpha}v_{\alpha}'^2\) - Final kinetic energy of the nucleus: \(E_N' = \frac{1}{2}m_Nv_N'^2\) Now, to find the percentage of the kinetic energy transferred from the alpha particle to the nucleus, we can calculate the difference between the initial and final kinetic energies of the alpha particle and compare it to the initial kinetic energy: Percentage of energy transferred = \(\frac{E_{\alpha} - E_{\alpha}'}{E_{\alpha}} \times 100\) After calculating the percentage of energy transferred, we will have the answer to the problem.

Key concepts

Conservation of Momentum

Momentum conservation is a fundamental principle in physics, especially when discussing collisions. This principle asserts that the total momentum of a closed system remains constant if it is not acted upon by external forces. Momentum, expressed as the product of an object's mass and its velocity, inherently links to motion and interactions.

In our scenario with an elastic alpha particle collision, we assume no external forces act on the particle and the nucleus. Thus, the total momentum before and after the collision must be equal. In mathematical terms, this is expressed as \(m_{\alpha}v_{\alpha} = m_{\alpha}v_{\alpha}' + m_Nv_N'\). This equation is crucial because it allows us to connect the initial and final states of the motion, providing one of the two needed equations to solve for the kinetic energy transfer in the collision.

Conservation of Kinetic Energy

The conservation of kinetic energy in an elastic collision is just as essential as the conservation of momentum. Elastic collisions are unique because they conserve both momentum and kinetic energy, which is the energy that an object possesses due to its motion.

The principle of kinetic energy conservation states that the total kinetic energy in an isolated system remains constant if there are no energy losses due to external work or conversion into other forms of energy. In an elastic collision between an alpha particle and a nucleus, expressed mathematically as \(\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \frac{1}{2}m_{\alpha}v_{\alpha}'^2 + \frac{1}{2}m_Nv_N'^2\), the kinetic energy before impact is equal to the kinetic energy after impact. This conservation law provides us with the second critical equation to describe the system's behavior fully and allows us to calculate the energy transfer once we have determined the final velocities.

Alpha Particle Collision

Alpha particles are helium nuclei and are a common example when studying nuclear collisions. In our example, an alpha particle collides with a much heavier stationary nucleus. The alpha particle has a mass of \(4.00 \textrm{u}\) (where \(u\) stands for atomic mass unit), and the nucleus has a mass of \(166 \textrm{u}\).

During the head-on elastic collision, the alpha particle transfers some of its kinetic energy to the nucleus. The collision is described as 'elastic' because it conserves both kinetic energy and momentum. The heavy mass of the nucleus compared to the alpha particle means that the alpha particle's final velocity will be significantly different from its initial velocity, which greatly influences the kinetic energy transfer. In nuclear physics, understanding the outcomes of such collisions is crucial for insight into nuclear reactions and decay processes.

Kinetic Energy Transfer

Kinetic energy transfer during a collision refers to the energy moved from one object to another as a result of the collision. In the context of our alpha particle and nucleus collision, the energy transfer is computed by finding the difference in the alpha particle's kinetic energy before and after the collision.

Calculating this energy transfer involves solving the equations for conservation of momentum and kinetic energy, yielding the final velocities of both the alpha particle and the nucleus. These velocities are then used to determine the individual kinetic energies post-collision. The percentage of kinetic energy transferred to the nucleus is particularly important because it reveals how much of the initial energy of the moving alpha particle was transferred to the previously stationary nucleus. In practical applications, understanding kinetic energy transfer is crucial for areas like particle physics, where energy transfer measurements are pivotal in identifying particle properties and interactions.