Problem 55 - University Physics with Modern Physics

Question

One proposed proton decay is \(\mathrm{p}^{+} \rightarrow \mathrm{e}^{+}+\pi^{0}\) which violates both baryon and lepton number conservation, so the proton lifetime is expected to be very long. Suppose the proton half-life were \(1.0 \times 10^{18} \mathrm{y}\) . (a) Calculate the energy deposited per kilogram of body tissue (in rad) due to the decay of the protons in your body in one year. Model your body as consisting entirely of water. Only the two protons in the hydrogen atoms in each \(\mathrm{H}_{2} \mathrm{O}\) molecule would decay in the manner shown; do you see why? Assume that the \(\pi^{0}\) decays to two \(\gamma\) rays, that the positron annihilates with an electron, and that all the energy produced in the primary decay and these secondary decays remains in your body (b) Calculate the equivalent dose (in rem) assuming an RBE of 1.0 for all the radiation products, and compare with the 0.1 rem due to the natural background and the 5.0 -rem guidelinefor industrial workers. Based on your calculation, can the proton lifetime be as short as \(1.0 \times 10^{18} \mathrm{y} ?\)

Step-by-Step Solution

Answer

The proton lifetime cannot be as short as 1.0 x 10^{18} years, as 0.695 rem is unreasonably high for a feasible lifetime.

Step 1: Calculate Number of Protons

Assume the body is composed entirely of water. The molecular weight of water (H\(_2\)O) is 18 g/mol, with hydrogen contributing 2 g/mol. Therefore, the number of protons (contributed by hydrogen atoms) per kilogram is calculated as: \[ \frac{2 \text{ g/mol}}{18 \text{ g/mol}} \times \frac{1 \text{ kg}}{12 \times 10^{-24} \text{ g/proton}} \] simplifying gives about 6.7 x 10\(^{25}\) protons/kg.

Step 2: Determine Proton Decay Rate

Given the proton half-life of 1.0 x 10\(^{18}\) years, the decay constant \(\lambda\) is calculated using: \[ \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.0 \times 10^{18}\, \text{years}} \] Calculating \(\lambda\) gives approximately 6.93 x 10\(^{-19}\) per year.

Step 3: Calculate Total Decays per Year

Using the decay constant \(\lambda\), the number of decays per year is: \[ N \times \lambda \] Substituting the number of protons (6.7 x 10\(^{25}\)) and \(\lambda\), the total decays per kilogram per year equals approximately 4.64 x 10\(^{7}\) decays per year/kg.

Step 4: Energy Deposited in Tissue

Each proton decay releases approximately 938 MeV (energy from decay and subsequent interactions), equivalent to \(938 \times 10^6 \) eV \(\times 1.6 \times 10^{-19}\) J/eV. The total energy deposited per kg is: \[ 4.64 \times 10^7 \times 938 \times 1.6 \times 10^{-13} \text{ J} \] which totals about 6.95 x 10\(^{-3}\) Joules/kg.

Step 5: Convert Energy to Rad

Since 1 rad is equivalent to 0.01 Joules/kg, convert the energy to rad: \[ \frac{6.95 \times 10^{-3}}{0.01} = 0.695 \text{ rad} \] This is the dose received per kilogram of tissue.

Step 6: Calculate Equivalent Dose in Rem

Using the Relative Biological Effectiveness (RBE) of 1.0 for each radiation type involved, the equivalent dose in rem is the same as the energy dose in rad: \[ 0.695 \text{ rem} = 0.695 \text{ rad} \] Compare this with the environmental dose of 0.1 rem and industrial allowance of 5.0 rem, 0.695 rem is higher than environmental levels but within industrial guidelines.

Key Concepts

Baryon Number Conservation

In physics, baryon number conservation is a fundamental principle that asserts the total number of baryons remains constant in an isolated system. Baryons are subatomic particles made up of three quarks, such as protons and neutrons. For every reaction or decay process, the baryon number before and after the event remains the same. This principle helps explain why certain reactions or decays are not observed in nature.

Proton decay, if it occurs, would violate baryon number conservation because the proton (a baryon) would transform into particles that are not baryons, such as positrons and pions. Current theoretical models predict that proton decay would have an incredibly long half-life due to this violation, making it extremely rare, if it happens at all.

Lepton Number Conservation

Lepton number conservation is another crucial concept in particle physics. Leptons are a class of subatomic particles that includes electrons, muons, and neutrinos. Similar to baryon number conservation, lepton number conservation states that the total number of leptons in a closed system remains constant through any reaction or decay.

For the proton decay reaction \(\mathrm{p}^{+} \rightarrow \mathrm{e}^{+} + \pi^{0}\), the initial lepton number is zero because protons and pions are not leptons. However, the end products include a positron, which is a lepton, resulting in a lepton number difference, thus violating the conservation law. Such violations are a key reason why proton decay is not typically observed.

Half-Life Calculation

Half-life is a measure of the time it takes for half of a given substance to undergo a specific reaction or decay process. It provides insight into the stability and longevity of particles.

The half-life of a proton, should it decay, is hypothesized to be around \(1.0 \times 10^{18}\) years. You can calculate the decay constant \(\lambda\) with the equation \(\lambda = \frac{0.693}{T_{1/2}}\), where \(T_{1/2}\) is the half-life.

This value is used to find the rate at which protons decay over a given period. In practical applications, knowing the half-life allows scientists to compute how many particles will decay in a certain amount of time, which aids in calculating radiation doses from particle decay.

Radiation Dose

Radiation dose is a measure of the energy deposited by ionizing radiation in a target material, such as body tissue. It is usually measured in rads for traditional units, where 1 rad equals 0.01 Joules of energy per kilogram of material.

When calculating the radiation dose from proton decay, all energy released during decay is expected to remain within the body. For example, in the problem scenario, each proton decay releases about 938 MeV of energy. By multiplying this energy by the number of decays per year and converting to Joules, you determine the energy deposited per kilogram.

This value, once converted to rads, reflects the dose received, allowing for comparison against natural background radiation levels and industrial safety standards.

Relative Biological Effectiveness (RBE)

Relative Biological Effectiveness (RBE) is a factor used to account for the differing impacts of various types of ionizing radiation on living tissue. Not all radiation has the same biological effect, even if the energy deposited is equivalent.

RBE compares the biological effectiveness of different kinds of radiation to the reference radiation (typically X-rays or gamma rays). An RBE of 1.0, as assumed in the proton decay scenario, means that the radiation involved is as biologically effective as the reference radiation.

To compute the equivalent dose in rems, multiply the energy dose in rads by the RBE. In this case, since RBE is 1.0, the equivalent dose is equal to the energy dose in rads, making calculations straightforward. This helps in assessing potential health impacts relative to established safety guidelines.