Question

\(^2$$^3$$^8\)U decays spontaneously by \(\alpha\) emission to \(^2$$^3$$^4\)Th. Calculate (a) the total energy released by this process and (b) the recoil velocity of the \(^2$$^3$$^4\)Th nucleus. The atomic masses are 238.050788 u for \(^2$$^3$$^8\)U and 234.043601 u for \(^2$$^3$$^4\)Th.

Short answer

(a) 4.27 MeV energy is released. (b) Recoil velocity of thorium is approximately 0.0129c.

Step-by-step solution

Identify the Decay Process

The given problem involves alpha decay, where \(^{238}_{92}\text{U} \rightarrow \, ^{234}_{90}\text{Th} + \, ^{4}_{2}\alpha\). We must account for both the thorium and alpha particle in the decay.

Calculate the Mass Defect

The mass defect is the difference in mass between the parent nucleus and the sum of the masses of the decay products. We calculate it as:\[\Delta m = m(^{238}_{92}\text{U}) - \left( m(^{234}_{90}\text{Th}) + m(^{4}_{2}\alpha) \right)\]We know: \(m(^{238}_{92}\text{U}) = 238.050788\, \text{u}\), \(m(^{234}_{90}\text{Th}) = 234.043601\, \text{u}\),and \(m(^{4}_{2}\alpha) \approx 4.002603\, \text{u}\) (mass of the alpha particle).\[\Delta m = 238.050788\, \text{u} - (234.043601\, \text{u} + 4.002603\, \text{u})\]\[\Delta m = 238.050788 - 238.046204 = 0.004584\, \text{u}\]

Convert Mass Defect to Energy

Convert the mass defect into energy using Einstein's equation:\(E = \Delta m \times c^2\).Using the conversion factor \(1\, \text{u} = 931.5\, \text{MeV/c}^2\):\[E = 0.004584\, \text{u} \times 931.5\, \text{MeV/u} \approx 4.27\, \text{MeV}\]

Calculate Recoil Velocity of \(^{234}_{90}\text{Th}\)

Use conservation of momentum to find the recoil velocity of the thorium nucleus. Assume the alpha particle takes most of the decay energy since it's much lighter.The momentum of the alpha particle is \( p_{\alpha} = m_{\alpha} \cdot v_{\alpha} \), and recoil momentum of thorium is \( p_{\text{Th}} = m_{\text{Th}} \cdot v_{\text{Th}} \). Conservation of momentum gives \(m_{\alpha} \cdot v_{\alpha} = m_{\text{Th}} \cdot v_{\text{Th}}\).Ignoring relativistic effects, \(v_{\alpha} \approx \sqrt{\frac{2E}{m_{\alpha}}}\).Thus:\[v_{\alpha} \approx \sqrt{\frac{2 \times 4.27\, \text{MeV}}{4.002603\, \text{u} \times 931.5\, \text{MeV/u}}} \approx \sqrt{2.135/3.73} \approx 0.755c\]Using conservation of momentum:\[v_{\text{Th}} = \frac{m_{\alpha}}{m_{\text{Th}}} \times v_{\alpha} \approx \frac{4.002603}{234.043601} \times 0.755c \approx 0.0129c\]

Key concepts

Mass Defect

Mass defect is a key concept in nuclear physics that explains why the mass of a nucleus is not just the sum of the masses of its individual protons and neutrons. It is the difference between the mass of a parent nucleus and the combined mass of its decay products. This discrepancy arises due to the conversion of some mass into binding energy, which holds the nucleus together.

To calculate the mass defect for a uranium-238 nucleus undergoing alpha decay to become thorium-234, we subtract the masses of thorium and the alpha particle from the initial uranium mass:

• Uranium-238 mass = 238.050788 u
• Thorium-234 mass = 234.043601 u
• Alpha particle mass \(\approx\) 4.002603 u

The mass defect is then: \[ \Delta m = 238.050788 u - (234.043601 u + 4.002603 u) = 0.004584 u. \]This loss in mass is responsible for the energy release in the decay process.

Energy Release

The energy released during nuclear decay is a result of the mass defect, as mass is converted into energy per Einstein’s famous equation \(E = mc^2\). Energy helps overcome the forces that keep the nucleus stable, allowing decay to occur. In this exercise, we convert the mass defect into energy using a conversion factor, showing how much energy is liberated during the decay process.

First, we convert the mass defect into energy: \(1 \, \text{u} = 931.5 \, \text{MeV/u}\), so:\[ E = 0.004584 \, \text{u} \times 931.5 \, \text{MeV/u} \approx 4.27 \, \text{MeV}. \] This means that as uranium-238 undergoes alpha decay, approximately 4.27 MeV of energy is released. This energy manifests as kinetic energy in the alpha particle and the newly formed thorium nucleus.

Recoil Velocity

Recoil velocity refers to the speed at which the thorium nucleus moves following alpha decay. As the alpha particle is emitted, the thorium nucleus recoils in the opposite direction to conserve momentum. This is akin to a gun recoiling back when it fires a bullet.

Since the alpha particle takes most of the energy due to its smaller mass, its momentum can be calculated as:\[ v_{\alpha} \approx \sqrt{\frac{2E}{m_{\alpha}}}, \]leading to: \[ v_{\alpha} \approx \sqrt{\frac{2 \times 4.27 \, \text{MeV}}{4.002603 \, \text{u} \times 931.5 \, \text{MeV/u}}} \approx 0.755c. \]For the thorium nucleus, the recoil velocity \(v_{\text{Th}}\) can be calculated by:\[ v_{\text{Th}} = \frac{m_{\alpha}}{m_{\text{Th}}} \times v_{\alpha} \approx \frac{4.002603}{234.043601} \times 0.755c \approx 0.0129c. \]This calculation demonstrates that although the thorium nucleus has a slower velocity compared to the emitted alpha particle, it moves with a significant speed due to conservation of momentum.

Conservation of Momentum

In the context of nuclear decay, conservation of momentum is paramount. This principle dictates that the total momentum of a system remains constant if no external forces act upon it. When uranium-238 undergoes alpha decay, it splits into two parts: the alpha particle and the thorium nucleus, both moving in opposite directions.

The momentum before the decay is zero because the uranium nucleus is initially at rest. After decay:

• The alpha particle’s momentum: \(p_{\alpha} = m_{\alpha} \cdot v_{\alpha}\)
• The thorium’s recoil momentum: \(p_{\text{Th}} = m_{\text{Th}} \cdot v_{\text{Th}}\)

Since momentum is conserved, we equate: \[ m_{\alpha} \cdot v_{\alpha} = m_{\text{Th}} \cdot v_{\text{Th}}. \]Thus, the recoil of thorium is crucially linked to the speed and mass of the emitted alpha particle. It exemplifies how energy release gets distributed during a nuclear transformation, aligning with physical laws.