Problem 55 - University Physics with Modern Physics

Question

Proton Decay. Proton decay is a feature of some grand unification theories. One possible decay could be \(\mathbf{p}^{+} \rightarrow \mathbf{e}^{+}+\pi^{0}\) , 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 a 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 guideline for 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 \(1.0 \times 10^{18}\) years; the calculated dose is negligible compared to the background radiation.

Step 1: Calculate the number of protons in the body

First, we need to calculate the number of protons in the body. Assume that the body is entirely made of water and weighs 70 kg. The mass of one water molecule (H₂O) is approximately 18 g/mol, which is equivalent to 0.018 kg/mol. Thus, we calculate the number of moles of water in a 70 kg body using the formula:\[\text{Number of moles of water} = \frac{70 \text{ kg}}{0.018 \text{ kg/mol}} = 3889 \text{ mol}\]Since each water molecule contains 2 hydrogen atoms (and thus 2 protons), the total number of protons in the body is:\[\text{Total protons} = 3889 \times 2 \times 6.022 \times 10^{23} \text{ protons/mol}\]= \(4.69 \times 10^{27}\) protons.

Step 2: Find the decay rate of protons

Given the half-life \(t_{1/2}\) of a proton is \(1.0 \times 10^{18} \text{ years}\), we find the decay constant \(\lambda\) using:\[\lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{1.0 \times 10^{18} \text{ years}}\]Converting years to seconds (since the decay rate requires seconds), we have:\[1 \text{ year} = 3.156 \times 10^{7} \text{ seconds}\]Thus, \(\lambda = \frac{0.693}{1.0 \times 10^{18} \times 3.156 \times 10^{7} \text{ s}}= 2.2 \times 10^{-26} \text{ s}^{-1}\).

Step 3: Calculate decays per year

To find the number of proton decays per year, use the decay constant:\[\text{Decays per year} = \lambda \times \text{Total protons}= 2.2 \times 10^{-26} \times 4.69 \times 10^{27}= 1.03 \times 10^{2} \text{ decays/year}\].

Step 4: Determine energy released per decay

The decay \(\mathbf{p}^{+} \rightarrow \mathbf{e}^{+} + \pi^{0}\) releases a combined energy of approximately 135 MeV (mass-energy of \(\pi^0\)) and 1.02 MeV for the positron annihilation, totaling 136.02 MeV per decay. Converting to joules:\[1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J} \, , \, \text{Total energy in joules} = 136.02 \times 1.602 \times 10^{-13} \text{ J} = 2.18 \times 10^{-11} \text{ J/decay}\].

Step 5: Calculate the energy deposited in the body

Multiply the energy released per decay by the number of decays per year to find the total energy deposited:\[\text{Energy per year} = 2.18 \times 10^{-11} \times 1.03 \times 10^{2} = 2.25 \times 10^{-9} \, \text{J}\].Converted to rads (1 rad = 0.01 J/kg for water):\[\text{Dose in rad} = \frac{2.25 \times 10^{-9} \text{ J}}{70 \text{ kg} \times 0.01}\approx 3.21 \times 10^{-9} \text{ rad/year}\].

Step 6: Calculate the equivalent dose in rem

Using a Relative Biological Effectiveness (RBE) of 1.0, the equivalent dose (in rem) can be determined by:\[\text{Dose in rem} = \text{Dose in rad} \times \text{RBE} = 3.21 \times 10^{-9} \text{ rem/year}\].Comparing this with the natural background dose (0.1 rem/year) and safety guideline (5 rem/year), the dose from proton decay is negligible.

Key Concepts

Grand Unification Theories

Grand Unification Theories (GUTs) represent a bold step in physics aiming to unify the three fundamental forces of the Standard Model: electromagnetic, weak, and strong nuclear forces.
Unlike the previous Electroweak Theory which only unified electromagnetic and weak forces, GUTs propose a single framework where these forces emerge from a grand unified force.
By proposing a relationship between forces that seem distinct at low energies, GUTs aim to simplify the myriad of elementary particles and interactions into a more elegant and coherent whole. Proton Decay as Evidence
A hallmark of many GUTs is the prediction of proton decay, a phenomenon not yet observed experimentally but has profound implications if detected. This decay process fits within GUTs as it embodies the idea of forces and particles changing into one another through a unified force interaction. The anticipation of proton decay requires proton stability as the forces unify at high energies—energies that were present only near the beginning of the universe. If protons were to decay, it would offer compelling support for these theories.

Baryon and Lepton Number Conservation

Baryon and lepton number conservation are foundational principles in the current understanding of particle physics. They serve as 'rules' that typically govern interactions among particles.
Baryons and leptons are two classes of particles, each represented by a number in reactions that must remain the same before and after a reaction if the laws hold.
- **Baryon Number Conservation**: Baryons, such as protons and neutrons, have a baryon number of +1. Proton decay, as proposed by some GUTs, would defy this conservation because a lone proton (baryon) turning into leptons, zeroing the baryon number, would otherwise be impossible under strict conservation laws. - **Lepton Number Conservation**: Leptons, including electrons and neutrinos, have lepton number conservation similar to baryons. Should a proton decay into a positron and other products, lepton number conservation would also be breached, sparking interest as this can suggest new physics beyond the Standard Model. These conservation laws, while not absolute under GUT propositions, give credence to the possible transformations at higher energies underlying the particle interactions.

Radiation Dose Calculation

Calculating radiation dose is crucial in assessing the effects of radiation exposure on biological tissues. This involves understanding both the energy absorbed by tissues and the quality of the radiation. Basic Steps in Dose Calculation
- **Energy Deposition**: The radiation dose is first determined by calculating the total energy deposited within the body. This involves multiplying the energy per decay by the number of decays, translating to how much radiative energy is absorbed. - **Converting to Radiation Dose Units**: The absorbed energy is then expressed in rads, a unit for measuring absorbed radiation dose within tissue, where 1 rad equals 0.01 Joules per kilogram. - **Equivalent Dose**: To assess potential biological effect, the absorbed dose (in rads) is adjusted by the radiation quality factor or RBE, resulting in the equivalent dose measured in rems. This process quantifies the biological impact of radiation, critical in setting safety standards and investigating hypothetical scenarios such as proton decay.

Half-life of Proton

The half-life of a proton, defined as the time it takes for half of a sample of protons to decay, remains a speculative but crucial aspect of nuclear physics. It provides insight into the stability of matter. Protons are extraordinarily stable, with current experiments suggesting their half-life exceeds observable periods far surpassing the age of the universe.
This presumed stability reinforces their role as fundamental building blocks of matter. Theoretical Implications
The hypothesized half-life in GUTs could still be as short as about 10^{31} years or more. However, the short lifespan explored in specific hypotheses, like 10^{18} years, is still beyond current experimental detection capabilities. Understanding proton half-life feeds directly back into GUTs and the hypothesized proton decay, offering hints of the universe's formative energies and forces that modern particle accelerators aim to replicate.

Radiation Effects on Human Body

Radiation can have significant effects on the human body, depending on the dose absorbed and the type of radiation involved. Acute vs. Chronic Effects
- **Acute Radiation**: High levels of radiation absorbed in a short period can cause immediate detrimental health effects such as radiation sickness, tissue damage, and increased cancer risk due to DNA mutations. - **Chronic Exposure**: Low-level radiation over prolonged periods can lead to cumulative effects, incrementally raising the risk of developing long-term health complications. Proton Decay Hypothetical Effects
In the context of proton decay, estimated yields can surprisingly result in minimal absorbed doses. Practical examples suggest that the energy resulting from a hypothetical proton decay, while biologically negligible compared to natural radiation background, underscores the importance of understanding each source's cumulative impacts. Effective dose management is vital, spectral in scenarios involving theoretical physics implications that previously unconcerned us, like proton decay or similar phenomena from high-energy physics.