Problem 71 - Physics for Scientists and Engineers with Modern Physics

Question

(a) In \(\alpha\) decay of, say, a \({ }_{88}^{226}\) Ra nucleus, show that the nucleus carries away a fraction \(1 /\left(1+\frac{1}{4} A_{\mathrm{D}}\right)\) of the total energy available, where \(A_{\mathrm{D}}\) is the mass number of the daughter nucleus. [Hint: Use conservation of momentum as well as conservation of energy.] (b) Approximately what percentage of the energy available is thus carried off by the \(\alpha\) particle when \({ }_{88}^{226} \mathrm{Ra}\) decays?

Step-by-Step Solution

Answer

The alpha particle carries approximately 98.2% of the energy in Ra-226 decay.

Step 1: Understand the Process of Alpha Decay

Alpha decay involves the emission of an alpha particle (which is equivalent to a ucleus of helium: two protons and two neutrons, or ). This decay reduces the atomic mass of the parent nucleus by 4 and the atomic number by 2.

Step 2: Conservation of Momentum

In decay, the conservation of momentum implies that the momentum of the alpha particle and the daughter nucleus must be equal and opposite. Therefore, if we let be the mass of the alpha particle, be its velocity, the mass of the daughter nucleus, and its velocity, then:\[ m_{\alpha} v_{\alpha} = m_{\text{D}} v_{\text{D}} \]

Step 3: Conservation of Energy

The total kinetic energy available, , is shared between the alpha particle and the daughter nucleus. Using the velocity expressions from step 2, the energy conservation gives us:\[ E = \frac{1}{2} m_{\alpha} v_{\alpha}^2 + \frac{1}{2} m_{\text{D}} v_{\text{D}}^2 \] Substituting momentum from the previous step, \( v_{\text{D}} = \frac{m_{\alpha}}{m_{\text{D}}} v_{\alpha} \), we obtain:\[ E = \frac{1}{2} m_{\alpha} v_{\alpha}^2 \left( 1 + \frac{m_{\alpha}}{m_{\text{D}}} \right) \]

Step 4: Simplify the Expression

We simplify the equation for the fraction of energy carried by the alpha particle, assuming the mass numbers can approximate the masses:\[ \frac{m_{\alpha}}{m_{\text{D}}} \approx \frac{4}{A_{\text{D}}} \]Therefore, the fraction of total energy carried by the alpha particle is:\[ \text{Fraction} = \frac{1}{1 + \frac{1}{4} A_{\text{D}}} \]

Step 5: Calculate for Radium-226 Decay

For the decay of , the daughter nucleus is with mass number 222:\[ A_{\text{D}} = 222 \] Substituting into the energy fraction formula gives:\[ \text{Fraction} = \frac{1}{1 + \frac{1}{4} } \approx \frac{1}{56.5} \]

Step 6: Final Calculation and Interpretation

Calculating approximately:\[ \frac{1}{56.5} \approx 0.982 \] This tells us that approximately 98.2% of the available energy is carried off by the alpha particle.

Key Concepts

Nuclear Physics

Nuclear physics is a fascinating branch of physics that studies the components and behavior of atomic nuclei. One phenomenon of interest within this field is radioactivity, which includes the process known as alpha decay. In alpha decay, a parent nucleus emits an alpha particle, which is a cluster of two protons and two neutrons. This leads to a reduction in both the atomic mass and atomic number of the original nucleus.
For example, when a radium-226 ( _{88}^{226} Ra ) nucleus undergoes alpha decay, it transforms into a radon-222 nucleus. The process is spontaneous and results in a more stable daughter nucleus. Alpha decay is a common decay mode among heavy elements and is crucial for understanding nuclear stability and radioactivity.
Studying such transformations helps us understand how elements change over time, influencing both the earth’s chemistry and biological systems. It also lays groundwork for technological applications such as nuclear energy and medical treatments.

Conservation of Energy

The principle of conservation of energy is fundamental in physics and asserts that energy cannot be created or destroyed, only transformed. In the context of alpha decay, this principle is crucial for calculating the distribution of energy between the particles involved.
When a nucleus decays by emitting an alpha particle, the conservation of energy ensures that the total kinetic energy before and after the decay remains constant. Mathematically, this can be expressed as:\[ E = \frac{1}{2} m_{\alpha} v_{\alpha}^2 + \frac{1}{2} m_{\text{D}} v_{\text{D}}^2 \] Here, \( E \) represents the total kinetic energy available from the decay, \( m_{\alpha} \) and \( v_{\alpha} \) are the mass and velocity of the alpha particle, and \( m_{\text{D}} \) and \( v_{\text{D}} \) are the mass and velocity of the daughter nucleus.
This relationship helps in determining how energy is partitioned, revealing that the alpha particle carries away most of the kinetic energy given its relatively small mass and high velocity compared to the daughter nucleus. Such calculations are essential in nuclear physics for predicting the behavior of decaying materials.

Conservation of Momentum

The conservation of momentum is another fundamental concept in physics which is particularly applicable in scenarios involving motion. According to this principle, the total momentum of a closed system must remain constant if no external forces are acting.
In the case of alpha decay, this means the momentum of the emitted alpha particle and the daughter nucleus must be equal and opposite. This can be expressed with the equation:\[ m_{\alpha} v_{\alpha} = m_{\text{D}} v_{\text{D}} \]Here, \( m_{\alpha} \) and \( v_{\alpha} \) refer to the mass and velocity of the alpha particle, while \( m_{\text{D}} \) and \( v_{\text{D}} \) refer to the same for the daughter nucleus.
This equation allows us to relate the velocities of the alpha particle and the daughter nucleus, which is critical for deriving energy relationships during the decay process. By knowing one velocity, the other can be calculated, enabling predictions about how momentum and energy are shared in nuclear decay processes. Understanding this concept is vital for comprehending the physical behavior of particles at the atomic scale.