Problem 75 - University Physics with Modern Physics

Question

The nucleus of \(^{214} \mathrm{Po}\) decays radioactively by emitting an alpha particle (mass \(6.65 \times 10^{-27} \mathrm{kg} )\) with kinetic energy \(1.23 \times\) \(10^{-12} \mathrm{J},\) as measured in the laboratory reference frame. Assuming that the Po was initially at rest in this frame, find the recoil velocity the nucleus that remains after the decay.

Step-by-Step Solution

Answer

The recoil velocity of the nucleus is approximately \(-9.19 \times 10^{5} \text{ m/s}\).

Step 1: Identify the Conservation Law

The key concept here is the conservation of momentum. Because no external forces act on the system, the total momentum before the decay must equal the total momentum after the decay. Before the decay, the momentum is zero since the nucleus is at rest.

Step 2: Calculate Alpha Particle Momentum

The momentum of the emitted alpha particle is given by its momentum formula, \( p = \sqrt{2mK} \), where \( m \) is the mass of the alpha particle and \( K \) is its kinetic energy. Substitute \( m = 6.65 \times 10^{-27} \text{ kg} \) and \( K = 1.23 \times 10^{-12} \text{ J} \):\[ p = \sqrt{2 \times 6.65 \times 10^{-27} \times 1.23 \times 10^{-12}} \]

Step 3: Momentum Balance Equation

According to conservation of momentum, the recoil momentum of the polonium nucleus must be equal and opposite to the momentum of the emitted alpha particle. Let \( M \) be the mass of the recoiling nucleus and \( v_r \) its recoil velocity. The equation is:\[ -p = M \cdot v_r \].Here, the negative sign indicates that the momentum direction of the two particles is opposite.

Step 4: Solve for Recoil Velocity

Isolate \( v_r \) in the momentum equation:\[ v_r = \frac{-p}{M} \].Given that the nucleus is significantly heavier (mass \( M \approx 214 \times 1.67 \times 10^{-27} \text{ kg} \)), use this to calculate \( v_r \) once you have \( p \) from Step 2.

Step 5: Perform the Calculation

First, compute \( p \):\[ p = \sqrt{2 \times 6.65 \times 10^{-27} \times 1.23 \times 10^{-12}} \approx 3.28 \times 10^{-19} \text{ kg} \cdot \text{m/s} \].Now calculate \( M \):\[ M \approx 214 \times 1.67 \times 10^{-27} = 3.57 \times 10^{-25} \text{ kg} \].Finally, calculate the recoil velocity:\[ v_r = \frac{-3.28 \times 10^{-19}}{3.57 \times 10^{-25}} \approx -9.19 \times 10^{5} \text{ m/s} \].

Key Concepts

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable nucleus transforms into a more stable one by ejecting an alpha particle. This process reduces the atomic mass of the original nucleus by four units and its atomic number by two units because an alpha particle consists of two protons and two neutrons, which makes it essentially a helium-4 nucleus.

Alpha decay is a natural way for heavy elements, like polonium, to lose mass and increase stability. It is often observed in elements with a high atomic number because the nuclear forces inside the atom struggle to hold such a large nucleus together.

• The emitted alpha particle is conserved in energy and momentum processes, making it pivotal in radioactive decay analysis.
• Despite being emitted with high energy and speed, alpha particles have a relatively low penetration power due to their large mass, meaning they can be easily stopped by a few centimeters of air or a sheet of paper.

Understanding alpha decay not only allows the calculation of particle energy and resultant velocities but also offers insights into nuclear stability and elemental transformations.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. In the context of particles, it is essential to understand how the energy of motion is expressed and utilized. The kinetic energy of a body can be simply calculated using the formula: \[ K = \frac{1}{2}mv^2 \] where \( m \) is the mass, and \( v \) is the velocity of the particle.

For the alpha particle mentioned in our problem, the given kinetic energy is a critical piece of information. It determines how fast the particle moves as it is emitted from the radioactive nucleus.

• Kinetic energy provides insights on the dynamics of radioactive decay, particularly the energy distribution after a decay event.
• When an alpha particle is emitted, the portion of the kinetic energy it carries away reduces the energy of the original nucleus, influencing its resultant state.
• By measuring kinetic energy, scientists can infer properties about the decay such as the force of ejection and the conditions inside the nucleus prior to the decay.

It serves as a conserved quantity across many physical transformations, thereby acting as a universal language in physics to describe and analyze motion.

Recoil Velocity

Recoil velocity is a result of the conservation of momentum principle, which in this case implies that the entirety of a system's momentum must be preserved through a process like alpha decay. When a nucleus releases an alpha particle, both the particle and what remains of the nucleus acquire velocity in opposite directions.

Since initially, the nucleus was at rest, the momentum of the system was zero. Post-decay, the momentum of the alpha particle must equal the momentum of the recoiling nucleus, hence their velocities are linked inversely by their respective masses.

• The formula used is derived from the momentum equation: \[ M \cdot v_r = -p \], where \( M \) is the mass of the recoiling nucleus and \( v_r \) is its speed.
• The negative sign in the velocity indicates opposite directionality between the alpha particle and the remaining nucleus.
• Recoil velocity is vital in nuclear physics to understand the effects and consequences of particle ejection, offering insights into the interactions within atomic structures.

This concept is widely applicable in both theoretical studies and practical applications such as nuclear reactors or radiation safety assessments.