Question

The nucleus of radioactive thorium -228 , with a mass of about \(3.78 \cdot 10^{-25} \mathrm{~kg},\) is known to decay by emitting an alpha particle with a mass of about \(6.64 \cdot 10^{-27} \mathrm{~kg} .\) If the alpha particle is emitted with a speed of \(1.80 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\), what is the recoil speed of the remaining nucleus (which is the nucleus of a radon atom)?

Short answer

Answer: The recoil speed of the remaining nucleus is approximately \(3.2 \cdot 10^5\mathrm{~m/s}\).

Step-by-step solution

Identify the principle to use

We will use the conservation of momentum principle, which states that the total momentum before a process is equal to the total momentum after the process.

Set up the momentum conservation equation

Let's denote the mass of the thorium nucleus as \(m_{Th}\), the mass of the alpha particle as \(m_\alpha\), the recoil speed of the remaining nucleus as \(v_{recoil}\), and the speed of the alpha particle as \(v_\alpha\). The conservation of momentum equation can be written as: \(m_{Th} \cdot 0 = m_\alpha \cdot v_\alpha + m_{Rn} \cdot v_{recoil}\) Note that the initial velocity of thorium nucleus is 0 (before decay).

Calculate the mass of the remaining nucleus

The mass of the remaining nucleus can be found by substracting the mass of the alpha particle from the mass of the thorium nucleus: \(m_{Rn} = m_{Th} - m_\alpha = 3.78 \cdot 10^{-25} kg - 6.64 \cdot 10^{-27} kg = 3.7136 \cdot 10^{-25} kg\)

Rearrange the equation to find the recoil speed

We need to find \(v_{recoil}\). Rearranging the equation from step 2: \(v_{recoil} = \frac{m_\alpha \cdot v_\alpha}{m_{Rn}}\)

Calculate the recoil speed

Substitute the values for \(m_\alpha\), \(v_\alpha\), and \(m_{Rn}\) into the equation to find \(v_{recoil}\): \(v_{recoil} = \frac{(6.64 \cdot 10^{-27} kg) \cdot (1.80 \cdot 10^{7} m/s)}{3.7136 \cdot 10^{-25} kg} \approx 3.2 \cdot 10^{5} m/s\) The recoil speed of the remaining nucleus (the radon atom) after the decay is approximately \(3.2 \cdot 10^5 \mathrm{~m/s}\).

Key concepts

Nuclear Decay

Nuclear decay represents the process by which an unstable atomic nucleus loses its energy. This occurs naturally in radioactive elements, like thorium-228. During decay, particles are emitted from the nucleus, and this can happen in various forms, including alpha decay, beta decay, or gamma decay. Alpha decay, specifically, involves the release of an alpha particle, which is essentially a helium nucleus consisting of two protons and two neutrons.

In this context, the thorium-228 nucleus undergoes such a transformation to reach a more stable state, ejecting an alpha particle in the process. The initial state of the thorium nucleus is typically stationary when it comes to its macroscopic motion, so we consider its initial speed to be zero in calculations related to its decay.

Radioactive Thorium-228

Thorium-228 is a radioactive isotope of thorium, and it's known for its alpha particle radioactive decay process. This isotope has a certain mass that can be used to calculate different properties when it decays. As an alpha emitter, when thorium-228 decays, it transforms into another element, in this case, a radon isotope (Rn) which is also radioactive. The mass of the thorium-228 nucleus plays a crucial role in determining the energy and speed with which the emitted alpha particles leave the nucleus.

Alpha Particle Emission

In the alpha particle emission process, an alpha particle, which is a type of ionizing radiation, is ejected from the nucleus of a radioactive atom. Alpha particles are relatively large, thus, they have a significant amount of kinetic energy upon emission.

After an alpha particle is emitted, there are two separate entities moving away from each other: the alpha particle itself and the remaining daughter nucleus (in our case, a radon atom). The speed of the ejected alpha particle is a crucial variable that determines the recoil speed of the daughter nucleus. Exercise-related calculations take the speed into account alongside the respective masses to uphold the principle of conservation of momentum.

Recoil Speed Calculation

The recoil speed of the remaining nucleus after alpha particle emission is calculated using the law of conservation of momentum. The premise is that the total momentum before the emission (of the thorium nucleus) must equal the total momentum after the event, comprising the momentum of both the alpha particle and the daughter nucleus (radon).

Since the thorium nucleus is initially at rest, its initial momentum is zero. After decay, the momentum is shared between the alpha particle and the newly formed radon nucleus. By rearranging the conservation of momentum equation, we can solve for the recoil speed of the radon nucleus. This recoil speed is significant because it quantifies the kinetic energy imparted to the daughter nucleus due to the decay process, providing insights into both the energy transformation and the physical dynamics at play during radioactive decay.