To solve the problem of finding the wavelength of each photon emitted during the annihilation of a positron and an electron, we can follow these steps:
### Step 1: Calculate the total energy before annihilation
The total energy of the system before annihilation consists of the rest mass energy and the kinetic energy of both the positron and the electron.
- **Rest mass energy of an electron (E₀)** = 0.51 MeV
- **Kinetic energy of positron (KE₁)** = 1 MeV
- **Kinetic energy of electron (KE₂)** = 1 MeV
The total energy (E_total) can be calculated as:
\[
E_{\text{total}} = E_{\text{rest}} + KE_{\text{positron}} + KE_{\text{electron}}
\]
\[
E_{\text{total}} = 0.51 \, \text{MeV} + 1 \, \text{MeV} + 0.51 \, \text{MeV} + 1 \, \text{MeV} = 3.02 \, \text{MeV}
\]
### Step 2: Determine the energy of each photon
During the annihilation process, the total energy is converted into two gamma photons. Therefore, the energy of each photon (E_photon) is:
\[
E_{\text{photon}} = \frac{E_{\text{total}}}{2} = \frac{3.02 \, \text{MeV}}{2} = 1.51 \, \text{MeV}
\]
### Step 3: Use the energy-wavelength relationship
The energy of a photon can also be expressed in terms of its wavelength (λ) using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \(h\) is Planck's constant (\(4.135667696 \times 10^{-15} \, \text{eV s}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))
We will rearrange this formula to find the wavelength:
\[
\lambda = \frac{hc}{E}
\]
### Step 4: Convert energy to joules
First, we need to convert the energy from MeV to joules.
1 MeV = \(1.6 \times 10^{-13}\) Joules, so:
\[
E_{\text{photon}} = 1.51 \, \text{MeV} = 1.51 \times 1.6 \times 10^{-13} \, \text{J} = 2.416 \times 10^{-13} \, \text{J}
\]
### Step 5: Calculate the wavelength
Now we can substitute the values of \(h\), \(c\), and \(E_{\text{photon}}\) into the wavelength equation:
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2.416 \times 10^{-13} \, \text{J}}
\]
Calculating this gives:
\[
\lambda \approx \frac{1.9878 \times 10^{-25}}{2.416 \times 10^{-13}} \approx 8.23 \times 10^{-14} \, \text{m}
\]
### Step 6: Convert to nanometers
To convert meters to nanometers:
\[
\lambda \approx 8.23 \times 10^{-14} \, \text{m} = 8.23 \times 10^{-5} \, \text{nm}
\]
### Final Answer
The wavelength of each photon emitted during the annihilation is approximately:
\[
\lambda \approx 8.23 \times 10^{-14} \, \text{m} \text{ or } 0.0823 \, \text{nm}
\]
