Problem 6.178 — Positron Emission Tomography (PET)

Problem Statement

Problem 6.178 — Positron Emission Tomography (PET)

Given Information

• All quantities, constants, and constraints stated in the problem above
• Physical constants used as needed (see Concepts section)

Physical Concepts & Formulas

This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.

• Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
• State the mathematical form of that law as it applies here
• Check dimensions at every step: both sides of an equation must have the same units

Step-by-Step Solution

Problem Statement

Problem 6.178 — Positron Emission Tomography (PET)

Given Information

• $= 2m$
• $E_{\gamma1} = E_{\gamma2} = m_ec^2 = 0.511 \text{ MeV}$

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

• See the step-by-step solution for the specific equations applied.
• All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: When a positron (nearly at rest) annihilates with an electron (at rest), two photons are produced. By conservation of energy and momentum:

Step 2 — Apply the relevant physical law or equation: Energy conservation: Total energy $= 2m_ec^2 = 2\times0.511 = 1.022$ MeV, split equally:

Step 3 — Solve algebraically for the unknown: $$E_{\gamma1} = E_{\gamma2} = m_ec^2 = 0.511 \text{ MeV}$$

Step 4 — Substitute numerical values with units: Momentum conservation: Initial momentum $\approx 0$, so two photons must travel in opposite directions:

Step 5 — Compute and check the result: $$\theta = 180°$$

Step 6: The two 511 keV gamma rays emerge in exactly opposite directions, allowing PET cameras (detector rings) to reconstruct the 3D location of the annihilation event (and thus the $^{18}$F tracer).

Worked Calculation

$$E_{\gamma1} = E_{\gamma2} = m_ec^2 = 0.511 \text{ MeV}$$

$$\theta = 180°$$

When a positron (nearly at rest) annihilates with an electron (at rest), two photons are produced. By conservation of energy and momentum:

Energy conservation: Total energy $= 2m_ec^2 = 2\times0.511 = 1.022$ MeV, split equally:

$$E_{\gamma1} = E_{\gamma2} = m_ec^2 = 0.511 \text{ MeV}$$

Momentum conservation: Initial momentum $\approx 0$, so two photons must travel in opposite directions:

$$\theta = 180°$$

The two 511 keV gamma rays emerge in exactly opposite directions, allowing PET cameras (detector rings) to reconstruct the 3D location of the annihilation event (and thus the $^{18}$F tracer).

Answers: Each photon has $E = 511$ keV; they emerge at $180°$ to each other.

Answer

$$\theta = 180°$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.

Worked Calculation

Substituting all given numerical values with their units into the derived formula:

$$\text{Numerical result} = \text{given expression substituted with values}$$

Answer

$$\boxed{\theta = 180°}$$

Physical Interpretation

The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.