Question

(a) Calculate the energy released when a U-238 isotope decays to Th-234. The atomic masses are as follows: U-238: 238.05078 amu; Th-234: 234.03596 amu; and He-4: 4.002603 amu. (b) The energy released in part (a) is transformed into the kinetic energy of the recoiling Th-234 nucleus and the $\alpha$ particle. Which of the two will move away faster? Explain.

Short answer

(a) Calculate the energy with the mass defect. (b) The $\alpha$ particle moves faster as it is lighter.

Step-by-step solution

Calculate Mass Defect

The decay of U-238 to Th-234 involves the emission of an alpha particle (He-4). The mass defect $\mathrm{\Delta }m$ is calculated as follows: $$\mathrm{\Delta }m=m_{\text{U-238}}-(m_{\text{Th-234}}+m_{\text{He-4}})$$ Substitute the given masses: $$\mathrm{\Delta }m=238.05078\text{amu}-(234.03596\text{amu}+4.002603\text{amu})$$ Calculate the mass defect.

Compute the Energy Released

The energy released $E$ from the decay is related to the mass defect through Einstein's equation $E=\mathrm{\Delta }m{c}^2$, where $c$ is the speed of light $3\times {10}^8\text{m/s}$. First, express the mass defect in kg using the conversion $1\text{amu}=1.660539\times {10}^{-27}\text{kg}$. Then, substitute $\mathrm{\Delta }m$ into the equation $E=\mathrm{\Delta }m{c}^2$ and calculate the energy in joules.

Determine Speeds of Th-234 and \alpha Particle

Since both the Th-234 nucleus and the $\alpha$ particle share the released energy as kinetic energy, use conservation of momentum to compare their speeds. Given that the momentum $p$ is $p=mv$, and both particles have equal and opposite momenta, solve for their velocities: $m_{\text{Th}}{v}_{\text{Th}}=m_{\alpha }{v}_{\alpha }$. Because $\alpha$ has a smaller mass, it will have a larger velocity compared to Th-234. Calculate and compare the two velocities based on the momentum formula.

Key concepts

Mass Defect

When U-238 undergoes decay to form Th-234, an alpha particle is emitted. This process involves a small difference in mass, known as the mass defect. The mass defect is simply the discrepancy between the original mass of the U-238 and the combined mass of the decay products, Th-234 and the alpha particle (He-4). This can be calculated using the formula:

• $\mathrm{\Delta }m=m_{\text{U-238}}-(m_{\text{Th-234}}+m_{\text{He-4}})$

Substituting in the atomic mass units (amu) provided, you'll find the mass defect directly. This step is crucial because it sets up the subsequent calculation of the energy released in the nuclear transformation.

Einstein's Equation

Einstein's Equation, famously known as $E=m{c}^2$, provides the bridge between mass and energy in the context of nuclear decay. Once the mass defect $\mathrm{\Delta }m$ is known, this equation allows us to find the total energy released during the decay process.
Start by converting the mass defect from atomic mass units to kilograms, using the conversion factor $1\text{amu}=1.660539\times {10}^{-27}\text{kg}$.
Then apply the formula:

• $E=\mathrm{\Delta }m{c}^2$
• where $c$ is the speed of light, approximately $3\times {10}^8\text{m/s}$.

This tells you how much energy, in joules, is liberated by the decay. Understanding this conversion is key to grasping the scale of energy involved in nuclear reactions.

Conservation of Momentum

In nuclear processes like alpha decay, the conservation of momentum is a principal law governing particle movement.
As the U-238 decays into Th-234 and an alpha particle, the conservation of momentum ensures that the decay products recoil back and forth in a way that their momenta are equal and opposite.
The formula for momentum $p=mv$ applies here. Given the two particles have equal and opposite momentum post-decay, we can express this relation as:

• $m_{\text{Th}}{v}_{\text{Th}}=m_{\alpha }{v}_{\alpha }$

From this relationship, since the alpha particle has a significantly smaller mass compared to Th-234, it will achieve a higher velocity. This principle is pivotal for understanding kinetic distribution among decay products.

Alpha Decay

Alpha decay is a type of radioactive decay where an atomic nucleus releases an alpha particle, composed of two protons and two neutrons. This process decreases the atomic number by two and the mass number by four, transitioning the original element to a new one.
In the given exercise, uranium-238 decays to thorium-234, releasing an alpha particle in the process. This is a common method by which heavy elements, especially those beyond lead on the periodic table, undergo transmutation to become more stable.
Alpha particles themselves are helium nuclei and have a relatively low penetration ability but a strong ionizing power, impacting how we observe and harness nuclear energy safety.

Kinetic Energy Distribution

Upon decay, the energy release is manifested as kinetic energy in the decay particles: Th-234 and the alpha particle. The total kinetic energy will be shared between the two. The kinetic energy distribution depends on the mass and velocity of the Th-234 nucleus and the alpha particle. Small mass alpha particles generally attain higher velocities compared to the heavier Thorium-234 nucleus, as derived from momentum conservation.
By calculating the velocities using the momentum relationship, one can determine how much kinetic energy each particle possesses and thus, gauge their speeds. This energy distribution explains why, although both particles move, the lighter alpha particle moves much faster than its heavy counterpart.