Problem 74 - University Physics with Modern Physics

Question

The nucleus of radioactive thorium- 228 , with a mass of about \(3.8 \cdot 10^{-25} \mathrm{~kg}\), is known to decay by emitting an alpha particle with a mass of about \(6.68 \cdot 10^{-27} \mathrm{~kg} .\) If the alpha particle is emitted with a speed of \(1.8 \cdot 10^{7} \mathrm{~m} / \mathrm{s},\) what is the recoil speed of the remaining nucleus (which is the nucleus of a radon atom)?

Step-by-Step Solution

Answer

Answer: The recoil speed of the remaining nucleus (radon atom) is approximately -3.22 * 10^5 m/s.

Step 1: Identify the initial momentum

Before the decay, both the thorium nucleus and the alpha particle are at rest, hence their total momentum is zero.

Step 2: Identify the final momentum

After the decay, the alpha particle and the remaining nucleus (radon atom) have respective momenta: p_alpha = m_alpha * v_alpha and p_radon = m_radon * v_radon. The total momentum after the decay is the sum of these two momenta: p_total = p_alpha + p_radon.

Step 3: Apply the conservation of momentum principle

Since the total momentum remains constant, we have: p_initial = p_total, implying 0 = p_alpha + p_radon.

Step 4: Express the momentum of the radon nucleus

We are looking for the recoil speed (v_radon) of the radon nucleus. From step 3, we have p_radon = -p_alpha. We can express the momentum of the radon nucleus as: m_radon * v_radon = -m_alpha * v_alpha.

Step 5: Calculate the mass of the radon nucleus

Since we know the mass of the thorium nucleus and the alpha particle, we can calculate the mass of the radon nucleus. The mass of the radon nucleus is the mass of the thorium nucleus minus the mass of the alpha particle: m_radon = m_thorium - m_alpha = 3.8*10^{-25} kg - 6.68*10^{-27} kg = (3.8*10^{-25} - 6.68*10^{-27})kg = (3.7332) * 10^{-25} kg.

Step 6: Calculate the recoil speed of the radon nucleus

Finally, we can find the recoil speed of the radon nucleus. From step 4, we have: v_radon = -(m_alpha * v_alpha) / m_radon = - (6.68*10^{-27} kg * 1.8*10^7 m/s) / (3.7332 * 10^{-25} kg) ≈ - (1.2024*10^{-19} kg*m/s) / (3.7332 * 10^{-25} kg) ≈ -3.22 * 10^5 m/s. The recoil speed of the remaining nucleus (radon atom) is approximately -3.22 * 10^5 m/s. The negative sign indicates that the radon nucleus moves in the opposite direction to the emitted alpha particle.

Key Concepts

Conservation of Momentum

Imagine you're on a skating rink, standing still on your skates. If you throw a ball away from your body, you’ll start moving in the opposite direction. This happens due to a principle known as the conservation of momentum, which is a fundamental concept in physics stating that the total momentum of a closed system remains constant if no external forces act upon it.

In the context of nuclear decay, like in the case of thorium-228, this principle tells us that the momentum before and after the decay event must be equal. Since the decay happens spontaneously, and there are no external forces acting on the nucleus, the initial momentum of the thorium nucleus (before decay) and the combined momentum of the alpha particle and radon nucleus (after decay) must balance each other out.

To put it simply, if the alpha particle shoots out in one direction, the radon nucleus must recoil in the opposite direction to keep the total momentum at zero, which was its original value when the thorium nucleus was at rest. This is why we observe 'recoil' in the radon atom post-decay.

Alpha Particle Emission

During alpha particle emission, a nucleus expels an alpha particle, which consists of two protons and two neutrons, essentially a helium-4 nucleus. This process is one way an unstable nucleus can transform to reach a more stable state.

In our example with thorium-228, the emission of the alpha particle is essentially the nucleus spitting out a chunk of itself. This is a process governed by the quantum mechanics operating within the nucleus. Yet, what we witness on a macroscopic level, such as the speed at which the alpha particle is emitted, is a result of these fundamental quantum phenomena. After the emission of an alpha particle, the original nucleus becomes a new element, in this case, a radon nucleus.

Alpha particles are relatively massive and carry away a significant amount of energy, but due to their size, they are not very penetrating and can be stopped by a sheet of paper, or even the dead layer of skin on a person's body. This type of radioactivity is important in fields like nuclear medicine and radiometric dating.

Recoil Speed

While solving a problem on recoil speed, always remember that it's the speed at which the leftover nucleus moves as a result of conservation of momentum. In our exercise regarding thorium-228, once it emits an alpha particle, it turns into a radon nucleus and moves in the opposite direction to maintain the system's initial momentum.

The recoil speed calculation is straightforward but crucial. It involves determining the mass of the recoiling nucleus and the speed of the emitted alpha particle. The recoil speed can provide us insights about the energy levels involved in the decay process and is also a key factor in applications like predictive models for radioactive behaviors and the design of nuclear reactors.

Students often find problems involving recoil speed challenging because they deal with both concepts of conservation laws and nuclear physics. However, by breaking down the problem as seen in the step-by-step solution, it becomes manageable to calculate and understand the physics behind the speed at which a radon nucleus recoils when thorium-228 decays.