Problem 77 - University Physics with Modern Physics

Question

Assume the nucleus of a radon atom, \({ }^{222} \mathrm{Rn}\), has a mass of \(3.68 \cdot 10^{-25} \mathrm{~kg} .\) This radioactive nucleus decays by emitting an alpha particle with an energy of \(8.79 \cdot 10^{-13} \mathrm{~J}\). The mass of an alpha particle is \(6.65 \cdot 10^{-27} \mathrm{~kg}\). Assuming that the radon nucleus was initially at rest, what is the velocity of the nucleus that remains after the decay?

Step-by-Step Solution

Answer

The velocity of the remaining nucleus after the decay is approximately -3.35 x 10^5 m/s. The negative sign indicates that the remaining nucleus moves in the opposite direction of the emitted alpha particle.

Step 1: Calculate the velocity of the alpha particle

To find the velocity of the alpha particle, we can use the following energy equation based on the given energy and mass: \(E_\alpha = \frac{1}{2} m_\alpha v_\alpha^2\) Where \(E_\alpha\) is the energy of the alpha particle, \(m_\alpha\) is the mass of the alpha particle, and \(v_\alpha\) is the velocity of the alpha particle. We can solve the equation for \(v_\alpha\) as follows: \(v_\alpha = \sqrt{\frac{2E_\alpha}{m_\alpha}}\) Using the given values, \(E_\alpha = 8.79 \cdot 10^{-13} \mathrm{~J}\) and \(m_\alpha = 6.65 \cdot 10^{-27} \mathrm{~kg}\), we get: \(v_\alpha = \sqrt{\frac{2(8.79 \cdot 10^{-13})}{6.65 \cdot 10^{-27}}}\) \(v_\alpha \approx 1.83 \cdot 10^7 \mathrm{~m/s}\)

Step 2: Apply conservation of linear momentum

The linear momentum before and after decay should be equal. Initially, the radon nucleus is at rest, so the momentum is zero. Let's denote the remaining nucleus by \(R'\), then the conservation of linear momentum can be expressed as: \(m_\alpha v_\alpha + m_{R'} v_{R'} = 0\) We can rearrange the equation to find \(v_{R'}\): \(v_{R'} = -\frac{m_\alpha}{m_{R'}} v_\alpha\)

Step 3: Find the mass of the remaining nucleus R'

To find \(m_{R'}\), we need to subtract the mass of the alpha particle from the initial mass of the radon nucleus: \(m_{R'} = m_{\mathrm{Rn}} - m_\alpha\) Using the given masses, \(m_{\mathrm{Rn}} = 3.68 \cdot 10^{-25} \mathrm{~kg}\) and \(m_\alpha = 6.65 \cdot 10^{-27} \mathrm{~kg}\), we have: \(m_{R'} = 3.68 \cdot 10^{-25} - 6.65 \cdot 10^{-27} \approx 3.61 \cdot 10^{-25} \mathrm{~kg}\)

Step 4: Calculate the velocity of the remaining nucleus R'

Now substitute the values for \(m_\alpha\), \(m_{R'}\), and \(v_\alpha\) into the equation from Step 2: \(v_{R'} = -\frac{6.65 \cdot 10^{-27}}{3.61 \cdot 10^{-25}} \cdot 1.83 \cdot 10^7\) We get: \(v_{R'} \approx -3.35 \cdot 10^5 \mathrm{~m/s}\) So, the velocity of the remaining nucleus R' after the decay is approximately \(-3.35 \cdot 10^5 \mathrm{~m/s}\). The negative sign indicates that the remaining nucleus moves in the opposite direction of the emitted alpha particle.

Key Concepts

Alpha Particle Emission

In nuclear physics, alpha particle emission is a type of radioactive decay where an unstable nucleus emits an alpha particle, which consists of two protons and two neutrons bound together. This process essentially reduces the atomic number by two and the mass number by four, transforming the original element into a new element with a different atomic structure.

During alpha decay, energy is released in the form of kinetic energy imparted to the alpha particle and the daughter nucleus. The energy originates from the difference in binding energies between the original nucleus and the final products. Alpha particles are relatively heavy and carry a double positive charge; as a result, they travel a short distance within materials and can be stopped by barriers as thin as paper or even the outer layer of human skin.

Conservation of Linear Momentum

The principle of conservation of linear momentum states that if no external forces act on a system, the total linear momentum of the system remains constant. In the context of radioactive decay, such as the alpha particle emission, this concept is crucial. Before decay, the radon nucleus is at rest, meaning its momentum is zero. When the alpha particle is emitted, to maintain the total momentum of zero, the daughter nucleus must recoil with momentum equal in magnitude but opposite in direction to that of the alpha particle.

This recoil is a key point in understanding the movement caused by nuclear decay. Due to this conservation, we can calculate the velocity and direction of the remaining nucleus after alpha particle emission, by considering the mass and velocity of the alpha particle. It also demonstrates the isolated nature of the decay event, wherein external forces are negligible.

Radioactive Decay

Radioactive decay is a stochastic (random) process at the level of single atoms, whereby an unstable atomic nucleus loses energy by radiation. This comes in several forms, including alpha decay, beta decay, gamma radiation, and more. Each of these processes helps the nucleus move toward a more stable configuration. During decay, the original element can transform into a different element or isotope. The rate of decay for a particular isotopic species is characterized by its half-life—the time it takes for half of any given amount of the isotope to decay.

Radioactive decay is a natural, spontaneous process that can be predicted very precisely for large numbers of atoms, but not for individual ones. Understanding radioactive decay allows for various applications in fields such as medicine, archaeology, and energy production.

Kinetic Energy Calculation

Kinetic energy (KE) represents the energy that an object possesses due to its motion. For any object with mass m traveling at velocity v, the kinetic energy can be described by the equation: \[KE = \frac{1}{2} m v^2\].In the context of the sample exercise, this equation allows us to calculate the velocity of the alpha particle based on its kinetic energy and mass. Upon computing the velocity of the alpha particle, we can subsequently apply the conservation of momentum to determine the velocity of the daughter nucleus post-decay.

By carefully applying the kinetic energy formula and considering the masses involved, the velocities of particles resulting from nuclear reactions such as alpha decay can be computed with considerable accuracy. This helps to predict and understand the behavior of the particles after such events.