Okay, let's tackle this problem. So, the question is about the annihilation of high-energy gamma rays with CMB photons into electron-positron pairs. The goal is to find the energy of these gamma rays where their lifetimes in the universe are limited by this process. The average CMB photon energy is given as 10^-3 eV. The options are A through D with various orders of magnitude in GeV.

First, I need to recall what processes limit the lifetime of high-energy gamma rays in the universe. The main idea here is that gamma rays can interact with other particles through various interactions. One such interaction is pair production, where a gamma photon collides with another photon (in this case, a CMB photon) to produce an electron-positron pair. The energy required for this process depends on the energies of both photons involved.

The process γγ → e+e- is a well-known interaction in particle physics. The threshold for this reaction is when the combined energy of the two photons is at least 2m_e c², where m_e is the electron mass. Since the rest mass energy of an electron is about 0.511 MeV, the threshold energy is 1.022 MeV. However, in reality, this process is suppressed unless there's a way for the photons to have the necessary energy and also for them to have the right relative directions and energies.

But in this case, one of the photons is from the CMB radiation, which has an average energy of 1e-3 eV. The other photon is the high-energy gamma-ray photon we're considering. The key here is the energy required for the process. Let me think: for the pair production to happen, the sum of the energies of the two photons in the center-of-momentum frame must be at least 2m_e c². But maybe a better approach is to consider the energy in the lab frame.

Alternatively, there's a concept called the "inverse Compton scattering" or maybe "photon-photon threshold" for pair production. The condition for the process γγ → e+e- to occur is that the total energy in the center-of-mass frame is at least twice the electron mass energy. The threshold in the lab frame might be different, but perhaps we can approximate it.

Wait, the problem states that the gamma rays are the ones with high energy, and the CMB photon is the low-energy one. So, let's model this: the CMB photon has energy E_photon (given as 1e-3 eV), and the gamma-ray photon has energy E. The total energy available for the pair production is E + E_photon, but that's in the lab frame where one photon is moving and the other is at rest (or the other way around). However, pair production typically requires that the two photons have enough combined energy in their center-of-mass frame. So the threshold would be when the sum of their energies in the center-of-mass frame equals 2m_e c².

But the center-of-mass energy squared s is given by s = (E1 + E2)^2 - (p1c + p2c)^2. For photons, since E = pc, this becomes s = (E1 + E2)^2 - (E1^2 + E2^2 + 2E1E2 cosθ)/c^2. Wait, maybe that's getting too complicated. Alternatively, the threshold condition for pair production by two photons is that the sum of their energies in the center-of-mass frame must be at least twice the electron mass energy. So, the minimum condition is when the photons are moving in opposite directions (head-on) to maximize the available energy.

In the case of a high-energy gamma ray and a low-energy CMB photon, if they are moving in opposite directions, then their energies add up constructively. So the center-of-mass energy would be sqrt(2E_gamma * E_CMB * (1 + cosθ)), but if θ is 180 degrees (opposite directions), then cosθ = -1, so the term becomes 1 - 1 = 0, which gives zero. Wait, that can't be right. Wait, maybe I need to recall the correct formula.

The correct formula for the invariant s (the square of the center-of-mass energy) for two photons with energies E1 and E2 and angle θ between them is s = 2E1E2(1 + cosθ). So when the photons are moving in opposite directions (θ=180°), cosθ = -1, so s = 2E1E2(1 -1) = 0. That would mean that the center-of-mass energy is zero, which can't be right. Wait, that can't be correct because pair production is possible when the photons are moving in the same direction? No, that doesn't make sense either. Wait, maybe the formula is different.

Wait, perhaps I should think in terms of energy and momentum conservation. For the process γ + γ → e+ + e-, the total energy and momentum must be conserved. Let's consider the lab frame where one photon is moving and the other is at rest. Wait, but photons can't be at rest. So maybe the lab frame is where one photon is moving, and the other is moving in the opposite direction. Wait, this is getting confusing. Let me try to approach this differently.

The threshold condition for pair production by two photons is that the sum of their energies in the center-of-mass frame must be at least 2m_e c². The minimum energy required is when the photons are moving directly towards each other (head-on collision), which maximizes the center-of-mass energy. In that case, the center-of-mass energy E_cm is given by sqrt(s), where s is the square of the invariant mass.

For two photons with energies E1 and E2 moving in opposite directions, the invariant s is:

s = (E1 + E2)^2 - (p1c - p2c)^2

But since photons have p = E/c, this becomes:

s = (E1 + E2)^2 - (E1/c - (-E2/c))^2 c^2

Wait, no, the momenta are vectors. If the photons are moving in opposite directions, their momenta are in opposite directions. So p1 = E1/c in one direction, p2 = E2/c in the opposite direction. So the total momentum is p1 - p2 (if p1 is in +x and p2 is in -x). So the invariant s is:

s = (E1 + E2)^2 - (p1 c - p2 c)^2

= (E1 + E2)^2 - (E1 - E2)^2

= [E1² + 2E1E2 + E2²] - [E1² - 2E1E2 + E2²]

= 4E1E2

So s = 4E1E2, which means the center-of-mass energy is sqrt(4E1E2) = 2 sqrt(E1 E2). Therefore, the threshold condition is when 2 sqrt(E1 E2) ≥ 2 m_e c². Dividing both sides by 2:

sqrt(E1 E2) ≥ m_e c²

Squaring both sides:

E1 E2 ≥ (m_e c²)^2

So the product of the two photon energies must be at least (0.511 MeV)^2. Let's compute this:

m_e c² = 0.511 MeV ≈ 0.511e6 eV

So (0.511e6 eV)^2 = (0.511)^2 * 1e12 eV² ≈ 0.261 * 1e12 eV² = 2.61e11 eV²

Therefore, E1 * E2 ≥ 2.61e11 eV²

In our problem, E2 is the CMB photon energy, which is 1e-3 eV. So E1 * (1e-3 eV) ≥ 2.61e11 eV²

Solving for E1:

E1 ≥ (2.61e11 eV²) / (1e-3 eV) = 2.61e14 eV

Convert eV to GeV: 1 GeV = 1e9 eV, so 2.61e14 eV = 2.61e5 GeV

So the gamma-ray energy E1 must be at least approximately 2.61e5 GeV. Looking at the options, option C is 2.6e5 GeV, which matches this calculation.

Wait, but let me check the calculation again. The CMB photon energy is 1e-3 eV, so E2 = 1e-3 eV. The product E1*E2 must be at least (0.511 MeV)^2. Let's compute:

0.511 MeV = 511 keV = 511,000 eV. So squared is (511,000 eV)^2 = (5.11e5 eV)^2 = 2.61e11 (eV)^2. Then E1 = 2.61e11 / 1e-3 = 2.61e14 eV. 1e14 eV is 1e5 GeV (since 1 GeV = 1e9 eV). So 2.61e14 eV = 2.61e5 GeV. Yes, that's 261,000 GeV. Option D is 2.6e5 GeV, which is 260,000 GeV, close to our calculation. The slight difference is probably due to rounding the 0.511 MeV to 0.511 exactly, but the options are given in orders of magnitude. So the correct answer should be C.
</think>