A moving positron and electron both with kinetic energy 1 Me V annihilate with each other and emits two gamma photons. If the rest mass energy of an electron is 0.51 MeV, the wavelength of each photon is ?

Options

A. `5.1xx10^(-3)Å`
B. `10.2xx10^(-3)Å`
C. `8.2xx10^(-3)Å`
D. `6.2xx10^(-3)Å`

Answer

To solve the problem of finding the wavelength of each photon emitted when a positron and an electron annihilate, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Energy in the System:**
- The kinetic energy of both the positron and the electron is given as 1 MeV each.
- The rest mass energy of an electron (and positron, since they have the same mass) is given as 0.51 MeV.
- Therefore, the total energy (E_total) of the system before annihilation can be calculated as:
\[ E_{\text{total}} = \text{K.E. of positron} + \text{K.E. of electron} + \text{Rest mass energy of positron} + \text{Rest mass energy of electron} \]
\[ E_{\text{total}} = 1 \, \text{MeV} + 1 \, \text{MeV} + 0.51 \, \text{MeV} + 0.51 \, \text{MeV} = 3.02 \, \text{MeV} \]

2. **Determine the Energy of Each Photon:**
- When the positron and electron annihilate, they produce two gamma photons.
- The energy of each photon (E_photon) can be calculated by dividing the total energy by 2:
\[ E_{\text{photon}} = \frac{E_{\text{total}}}{2} = \frac{3.02 \, \text{MeV}}{2} = 1.51 \, \text{MeV} \]

3. **Convert Energy to Joules:**
- To find the wavelength, we need to convert the energy from MeV to Joules. The conversion factor is:
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
- Therefore, the energy in Joules is:
\[ E_{\text{photon}} = 1.51 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 2.416 \times 10^{-13} \, \text{J} \]

4. **Use the Energy-Wavelength Relation:**
- The energy of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
- Rearranging for wavelength gives:
\[ \lambda = \frac{hc}{E} \]
- Here, \( h \) (Planck's constant) is \( 6.63 \times 10^{-34} \, \text{J s} \) and \( c \) (speed of light) is \( 3 \times 10^8 \, \text{m/s} \).

5. **Calculate the Wavelength:**
- Plugging in the values:
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{2.416 \times 10^{-13} \, \text{J}} \]
- Performing the calculation:
\[ \lambda \approx \frac{1.989 \times 10^{-25}}{2.416 \times 10^{-13}} \approx 8.23 \times 10^{-3} \, \text{m} = 8.23 \, \text{pm} \text{ (picometers)} \]

### Final Answer:
The wavelength of each photon emitted during the annihilation is approximately \( 8.23 \, \text{pm} \) or \( 8.23 \times 10^{-3} \, \text{angstroms} \).