Problem 71 - Physics for Scientists and Engineers with Modern Physics

Question

(a) In \(\alpha\) decay of, say, a \(\frac{226}{88}Ra\) 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 \(\frac{226}{88}Ra\) decays?

Step-by-Step Solution

Answer

The \(\alpha\) particle carries approximately 1.77% of the total energy released in the decay.

Step 1: Understand the Decay Process

In an \(\alpha\) decay, the original nucleus (parent), \( ^{226}_{88} \text{Ra} \), transforms into a daughter nucleus and releases an \(\alpha\) particle. The \(\alpha\) particle is essentially a \( ^{4}_{2} \text{He} \) nucleus. Thus, for the decay of \( ^{226}_{88} \text{Ra} \), the daughter nucleus becomes \( ^{222}_{86} \text{Rn} \). Let's define \(A_{\text{D}} = 222\) for the daughter nucleus.

Step 2: Apply Momentum Conservation

Since momentum is conserved, the momentum of the \(\alpha\) particle \(p_{\alpha}\) and the daughter nucleus \(p_{\text{D}}\) are equal and opposite: \ p_{\alpha} = -p_{\text{D}}.\ The relation between energy \(E\) and momentum \(p\) for a non-relativistic particle is given by:\[ p = \sqrt{2mE}, \] where \(m\) is the mass of the particle.

Step 3: Apply Energy Conservation

The total energy released in the decay \(Q\) is shared between the kinetic energies of the \(\alpha\) particle \(E_{\alpha}\) and the daughter nucleus \(E_{\text{D}}\). Thus, we can write: \[ Q = E_{\alpha} + E_{\text{D}}. \] Using the momentum conservation equation, and equating energies, we can show:\[ \frac{E_{\alpha}}{E_{\text{D}}} = \frac{m_{\text{D}}}{m_{\alpha}}. \] Since \( m_{\text{D}} \approx A_{\text{D}} \) and \( m_{\alpha} \approx 4 \) (mass numbers), we solve to find the fraction of energy carried away by the \(\alpha\) particle.

Step 4: Calculate the Energy Fraction

Using the relation from Step 3, the fraction of energy carried by the \(\alpha\) particle is: \[ \frac{E_{\alpha}}{Q} = \frac{m_{\text{D}}}{m_{\alpha} + m_{\text{D}}} = \frac{A_{\text{D}}}{4 + A_{\text{D}}}. \]For the daughter nucleus of mass number \(A_{\text{D}}\), the fraction of total energy carried by the daughter nucleus is \ \( \frac{1}{1 + \frac{1}{4}A_{\text{D}}} \). This represents the fraction of energy carried by the \(\alpha\) particle.

Step 5: Specific Calculation for Ra-226

For \( ^{226}_{88} \text{Ra} \), \(A_{\text{D}} = 222\). Plug this into the fraction: \ \( \frac{1}{1 + \frac{1}{4} \times 222} \approx \frac{1}{56.5} \). \Thus, the \(\alpha\) particle carries away about \( \frac{1}{56.5} \) of the energy released.

Step 6: Calculate Percentage

To find the percentage of energy carried off by the \(\alpha\) particle, convert the fraction \(\frac{1}{56.5}\) into a percentage: \ \( \frac{1}{56.5} \times 100\% \approx 1.77\% \). Thus, approximately 1.77\% of the energy is carried by the \(\alpha\) particle.

Key Concepts

Conservation of Momentum

In the interesting world of nuclear physics, one of the key principles at play during alpha decay is the conservation of momentum. When a radiation source, such as the isotope \( ^{226}_{88} \text{Ra} \), undergoes alpha decay, it changes into a different element. This transformation involves the parent nucleus releasing an alpha particle, which is almost like ejecting a tiny ball into space. Momentum, which is a measure of motion, must remain constant before and after this event. This means that the momentum of the alpha particle must exactly counterbalance that of the remaining nucleus, known as the daughter nucleus.
As these two separate, the alpha particle and the daughter nucleus recoil in opposite directions. The mathematical relationship describing this balance is simple: \[ p_\alpha = -p_\text{D}, \]where \( p_\alpha \) represents the momentum of the alpha particle, and \( p_\text{D} \) is the momentum of the daughter nucleus. Despite the complexities of nuclear physics, this straightforward balancing act maintains the harmony of the universe at a subatomic level.

Conservation of Energy

Alongside momentum, energy is also conserved in the process of alpha decay. Consider the original energy available in the parent nucleus as a pot of gold. When the nucleus decays, this pot of gold is split between the alpha particle and the daughter nucleus as kinetic energy. The total energy \( Q \) released in the process is divided between these two particles, and is described by the equation:\[ Q = E_\alpha + E_\text{D}. \]Here, \( E_\alpha \) represents the kinetic energy of the alpha particle and \( E_\text{D} \) symbolizes the kinetic energy of the daughter nucleus.
The division of energy relates directly to the masses of the alpha particle and the daughter nucleus. With the application of the energy conservation principle and knowledge of their respective masses, the energy fraction can be established. Here, the alpha particle takes:\[ \frac{E_\alpha}{Q} = \frac{m_\text{D}}{m_\alpha + m_\text{D}} = \frac{A_\text{D}}{4 + A_\text{D}}, \]where \( A_\text{D} \) refers to the mass number of the daughter nucleus. By solving this formula, we realize how much energy the alpha particle steals away when it makes its escape, like a pirate with a treasure map.

Daughter Nucleus

In alpha decay, the term "daughter nucleus" stands for what remains of the original parent nucleus after it ejects an alpha particle. For example, when \( ^{226}_{88} \text{Ra} \) decays, it transmutes into \( ^{222}_{86} \text{Rn} \). This new element is the daughter nucleus. The term 'daughter' might evoke familial bonds, and indeed, they share a common origin, which is the parent nucleus. The daughter nucleus plays a crucial role in the balance of energy and momentum. It is as if after birth, it carries on part of the legacy, by holding the rest of the original energy not seized by the outgoing alpha particle. Its mass number, \( A_\text{D} \), influences how the energy is divided during decay, determining what fraction of energy is retained by the daughter itself.For a \( ^{226}_{88} \text{Ra} \) alpha decay event, the daughter nucleus \( ^{222}_{86} \text{Rn} \) plays its part in maintaining the delicate cosmic balances of momentum and energy, showing how each event has its own story of conservation and change.