Question

(a) Calculate the energy released when a \({ }^{238} \mathrm{U}\) isotope decays to \({ }^{234} \mathrm{Th} .\) The atomic masses are given by: \(^{238} \mathrm{U}: 238.0508 \mathrm{amu} ;{ }^{234} \mathrm{Th}: 234.0436 \mathrm{amu} ;{ }^{4} \mathrm{He}\) 4.0026 amu. (b) The energy released in (a) is transformed into the kinetic energy of the recoiling \({ }^{234} \mathrm{Th}\) nucleus and the \(\alpha\) particle. Which of the two will move away faster? Explain.

Short answer

The energy released during the decay is \(4.28 \, \text{MeV}\). The alpha particle (Helium-4 nucleus) moves away faster post-decay.

Step-by-step solution

Calculate Mass Deficit

The change in mass, which is often referred to as the 'mass deficit', results in energy release. Calculate it by subtracting the mass of the products (Thorium-234 and Helium-4) from the initial mass of Uranium-238. We get: \( \Delta m = m_{^{238}U} - (m_{^{234}Th} + m_{^{4}He}) = 238.0508 \, \text{amu} - (234.0436 \, \text{amu} + 4.0026 \, \text{amu}) = 0.0046 \, \text{amu} \)

Convert Mass Deficit to Energy

Using Einstein's relation \(E = \Delta mc^2\), where \(\Delta m\) is the mass difference and \(c\) is the speed of light. However, when working in atomic mass units, we use the conversion factor that \(1 \, \text{amu} = 931.5 \, \text{MeV}/c^2\). Thus, the energy released (E) is:\(E = \Delta m \times 931.5 \, \text{MeV}/c^2 = 0.0046 \, \text{amu} \times 931.5 \, \text{MeV}/c^2 = 4.28 \, \text{MeV}\)

Consider Momentum Conservation

The momentum of the system before decay equals the momentum of the system after decay. Before decay, the Uranium-238 atom was at rest, so its momentum was zero. After decay, the momentum of the Thorium-234 and the alpha particle must therefore also add up to zero due to the law of conservation of momentum. This results in an 'equal and opposite' momentum scenario, meaning the larger mass (Thorium-234) will have a smaller speed, while the smaller mass (alpha particle) will have a larger speed.

Key concepts

Mass Deficit

In nuclear decay, a phenomenon known as "mass deficit" is observed. This is where the mass of the original nucleus (in this example, Uranium-238) is greater than the sum of the masses of its decay products (Thorium-234 and Helium-4).

The deficit is calculated by subtracting the total mass of the decay products from the initial mass of the decaying nucleus:

• For Uranium-238 decay: \[ \Delta m = m_{^{238}U} - (m_{^{234}Th} + m_{^{4}He}) \]
• This results in a mass difference or deficit of 0.0046 amu.

This mass deficit is crucial because it directly ties into how much energy is released during the decay process. The smaller resulting mass indicates that some of the mass has been converted into energy, which we explore next.

Energy Conversion

The mass deficit we calculated isn't just wasted; it converts into energy, a principle famously described by Einstein's equation \( E = mc^2 \). This conversion is central to understanding nuclear reactions.

To convert the deficit into energy:

• First, recognize that in the context of nuclear physics, we often use atomic mass units (amu) and convert them into energy measured in Mega-electronvolts (MeV).
• The conversion factor is approximately 931.5 MeV per amu.
• Applying this factor, the energy release from our mass deficit is:\[E = \Delta m \times 931.5 \, \text{MeV}/c^2 = 0.0046 \, \text{amu} \times 931.5 \, \text{MeV}/c^2\]
• This results in an energy output of 4.28 MeV.

This energy release is part of what makes nuclear reactions so powerful and efficient.

Momentum Conservation

After the decay of Uranium-238, the momentum before and after the decay must remain balanced, based on the law of conservation of momentum. Initially, the uranium nucleus is at rest, meaning its momentum is zero. Once decay happens, two new particles emerge – the Thorium-234 nucleus and an alpha particle.

To preserve momentum:

• The total momentum of these particles must remain zero (the same as the initial state).
• This results in the condition that the momentum of Thorium-234 and the alpha particle must be equal in magnitude but opposite in direction.

Since momentum is the product of mass and velocity, and knowing that Thorium is much heavier than the alpha particle:

• The alpha particle, having a smaller mass, must move away at a higher speed to balance the momentum equation.
• Accordingly, Thorium-234 moves more slowly due to its larger mass.

Thus, the remarkable balance of momentum inherent in nuclear decay gives us insights into the dynamics of subatomic particles. It shows us how even the absence of movement initially retains its balance post-decay through the interplay of mass and velocity.