Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the wavelength of the resulting radiation?

Options

A. `lambda = (h)/(m_(0)c)`
B. `lambda = (2 h)/(m_(0)c^(2))`
C. `lambda = (h)/(2 m_(0)c^(2))`
D. None of these

Explanation

To solve the problem of finding the wavelength of the radiation resulting from the annihilation of a stationary electron and a stationary positron, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**: When an electron (e⁻) and a positron (e⁺) annihilate each other, they produce two photons (γ) that travel in opposite directions.

2. **Conservation of Energy**: The total energy before the annihilation must equal the total energy after the annihilation. The rest energy of the electron and positron can be expressed using Einstein's equation \(E = mc^2\).
- The rest mass energy of the electron (m₀) is \(E_{e^-} = m_0 c^2\).
- The rest mass energy of the positron is the same, \(E_{e^+} = m_0 c^2\).
- Therefore, the total energy before annihilation is:
\[ E_{\text{total}} = E_{e^-} + E_{e^+} = m_0 c^2 + m_0 c^2 = 2m_0 c^2 \]

3. **Energy of Photons**: After annihilation, the energy is carried away by two photons. The energy of each photon can be expressed as:
\[ E_{\gamma} = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon.

4. **Total Energy of Photons**: Since there are two photons, the total energy of the photons is:
\[ E_{\text{photons}} = 2E_{\gamma} = 2 \left(\frac{hc}{\lambda}\right) \]

5. **Setting Energies Equal**: According to the conservation of energy, we set the total energy before annihilation equal to the total energy after annihilation:
\[ 2m_0 c^2 = 2 \left(\frac{hc}{\lambda}\right) \]

6. **Simplifying the Equation**: We can simplify this equation by dividing both sides by 2:
\[ m_0 c^2 = \frac{hc}{\lambda} \]

7. **Solving for Wavelength**: Rearranging the equation to solve for \(\lambda\):
\[ \lambda = \frac{hc}{m_0 c^2} \]
Simplifying further, we get:
\[ \lambda = \frac{h}{m_0 c} \]

8. **Final Expression**: This is the wavelength of the resulting radiation from the annihilation of a stationary electron and positron.

### Final Answer:
\[ \lambda = \frac{h}{m_0 c} \]