Problem 52 - Fundamentals of Physics

Question

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays $${ }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C} \quad \text { and } \quad{ }^{223} \mathrm{Ra} \rightarrow{ }^{219} \mathrm{Rn}+{ }^{4} \mathrm{He}$$ Calculate the \(Q\) value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is \(30.0 \mathrm{MeV}\). What is the barrier height for \({ }^{14} \mathrm{C}\) emission? (Be careful about the nuclear radii.) The needed atomic masses are $$ \begin{aligned} &\begin{array}{llll} { }^{223} \mathrm{Ra} & 223.01850 \mathrm{u} & { }^{14} \mathrm{C} & 14.00324 \mathrm{u} \end{array}\\\ &{ }^{209} \mathrm{~Pb} \quad 208.98107 \mathrm{u} \quad{ }^{4} \mathrm{He} \quad 4.00260 \mathrm{u}\\\ &{ }^{219} \mathrm{Rn} \quad 219.00948 \mathrm{u} \end{aligned} $$

Step-by-Step Solution

Answer

a) 31.88 MeV, b) 6.40 MeV, both decays are possible. c) Barrier height for \({}^{14}C\) is approximately 155.9 MeV.

Step 1: Understand the Q Value Formula

The Q value of a nuclear reaction is given by the formula \( Q = \left( m_i - m_f \right) \times c^2 \), where \( m_i \) is the initial mass and \( m_f \) is the final mass of the products. It's important to know that \( c \) is the speed of light, but we will work in atomic mass units (u) to electronvolts (MeV). 1 u corresponds to 931.5 MeV.

Step 2: Calculate Q Value for Ra to Pb and C

For the first decay \( {}^{223} \mathrm{Ra} \rightarrow {}^{209} \mathrm{Pb} + {}^{14} \mathrm{C} \), the Q value is computed as follows:\[ m_i = m({}^{223} \mathrm{Ra}) = 223.01850 \, \text{u} \]\[ m_f = m({}^{209} \mathrm{Pb}) + m({}^{14} \mathrm{C}) = 208.98107 \, \text{u} + 14.00324 \, \text{u} = 222.98431 \, \text{u} \]\[ Q = (223.01850 - 222.98431) \, \text{u} \times 931.5 \frac{\mathrm{MeV}}{\mathrm{u}} \approx 31.88 \, \text{MeV} \]

Step 3: Check Energy Possibility for First Decay

If the Q value is positive, the decay is energetically possible. Since we calculated \( Q = 31.88 \, \text{MeV} \) for the first decay, it is energetically feasible.

Step 4: Calculate Q Value for Ra to Rn and He

For the second decay \( {}^{223} \mathrm{Ra} \rightarrow {}^{219} \mathrm{Rn} + {}^{4} \mathrm{He} \), the Q value is:\[ m_i = m({}^{223} \mathrm{Ra}) = 223.01850 \, \text{u} \]\[ m_f = m({}^{219} \mathrm{Rn}) + m({}^{4} \mathrm{He}) = 219.00948 \, \text{u} + 4.00260 \, \text{u} = 223.01208 \, \text{u} \]\[ Q = (223.01850 - 223.01208) \, \text{u} \times 931.5 \frac{\mathrm{MeV}}{\mathrm{u}} \approx 6.40 \, \text{MeV} \]

Step 5: Check Energy Possibility for Second Decay

As with the first decay, a positive Q value (6.40 MeV) indicates that the decay \( {}^{223} \mathrm{Ra} \rightarrow {}^{219} \mathrm{Rn} + {}^{4} \mathrm{He} \) is energetically possible.

Step 6: Understand Coulomb Barrier Concept

The Coulomb barrier is the energy barrier due to electrostatic interaction that the emitted particle must overcome to escape the nucleus. It's usually larger for larger emitted particles due to increased charge.

Step 7: Calculate Coulomb Barrier Height for C-14 Emission

Assuming the nuclear radius to be proportional to \( A^{1/3} \), the radius ratio between \( {}^{14} \mathrm{C} \) and \( {}^{4} \mathrm{He} \) is roughly \( (14/4)^{1/3} \approx 1.3195 \). The charge ratio is \( (6/2) = 3 \). Hence, approximate barrier height:\[ \frac{Z_{C}}{Z_{\alpha}} \times \left( \frac{R_{C}}{R_{\alpha}} \right)^{2} \times 30.0 \, \text{MeV} = 3 \times 1.32^2 \times 30.0 \, \text{MeV} \approx 155.9 \, \text{MeV} \]

Key Concepts

Q value calculation

In nuclear physics, the Q value helps us determine if a nuclear reaction or decay is energetically possible. It represents the energy released or absorbed during the reaction. The formula to calculate the Q value is:\[ Q = (m_i - m_f) \times c^2 \]Here, \(m_i\) is the mass of the initial nucleus, \(m_f\) is the total mass of the decay products, and \(c\) is the speed of light. However, we often use atomic mass units (u) and convert using the fact that 1 u equals 931.5 MeV.
To evaluate whether a reaction is possible, calculate the difference between the initial and final masses, then multiply by 931.5 MeV/u. A Q value greater than zero indicates that the reaction can occur, as it releases energy. This forms the basis of our approach to understanding nuclear decays and is crucial for checking energetic feasibility.

Coulomb barrier

The Coulomb barrier is an essential concept in nuclear physics, representing the energy threshold that a charged particle must overcome to escape the attractive forces of the nucleus. This energy barrier arises from the electrostatic force between the positively charged particles in the nucleus and the emitted particle, and it's more significant for particles with higher electric charge.
When thinking about the Coulomb barrier, remember:

• The higher the charge of the emitted particle, the higher the Coulomb barrier, as like charges repel each other.
• This barrier limits the likelihood of spontaneous emission or capture by another nucleus.
• If the particle lacks sufficient energy to surpass the barrier, it will not escape the nucleus.

Calculating the height of this barrier for a specific decay helps clarify whether the particle can naturally overcome these forces in nuclear reactions.

Alpha particle emission

Alpha particles are Helium nuclei consisting of 2 protons and 2 neutrons. They are relatively massive and positively charged. Alpha emission is a type of radioactive decay where these particles are spontaneously emitted from a nucleus. Key characteristics include:

• Alpha particles have a +2 charge, meaning they face a significant Coulomb barrier compared to smaller particles.
• They lose energy quickly due to their mass and charge, limiting their range in materials.
• This type of decay is common in heavy elements, like Uranium or Radium, where large nuclei become more stable by shedding excess nucleons.

Understanding alpha emissions allows us to explore various nuclear stability phenomena and energy release processes, essential for comprehending heavy element decay chains.

Atomic mass units

The atomic mass unit (amu or u) is a standard unit for expressing atomic and molecular masses. It is based on one-twelfth the mass of a carbon-12 atom. Using amu simplifies calculations in nuclear physics by providing a consistent mass scale. Important points about atomic mass units:

• One atomic mass unit is approximately equivalent to 1.66053906660 x 10^-27 kg.
• The conversion factor for energy calculations: 1 u equals 931.5 MeV.
• It allows scientists to perform calculations relating to nuclear binding energy, decay exercises, and reaction energetics uniformly.

In calculations, mass values are often expressed in atomic mass units, providing a convenient pathway to bridge between mass and energy computations critical in nuclear decay.

Nuclear radii

Nuclear radii describe the size of an atomic nucleus. This is crucial when considering nuclear reactions and decay because the nuclear radius affects the Coulomb barrier, especially when different nuclei are involved.To estimate a nuclear radius, we consider that it roughly scales with the atomic mass number, A, using:\[ R \approx R_0 \cdot A^{1/3} \]where \(R_0\) is approximately 1.2-1.3 femtometers.Key insights into nuclear radii:

• It helps calculate the spatial distribution of nucleons within the nucleus.
• Plays a role in determining collision probabilities and interaction likelihoods in nuclear reactions.
• Affects the calculations of forces, such as the magnitude of the Coulomb barrier, as it relates to the proximity of neighboring nuclei.

By appreciating these concepts, we delve deeper into the dynamics of nuclear stability and reactions, illustrating the crucial role that nuclear dimensions play in physics.