The positron is a fundamental particle with the same mass as that of the electron and with a charge equal to that of an electron but of opposite sign. When a positron and an electron collide, they may annihilate each other. The energy corresponding to their mass appears in two photons of equal energy. Find the wavelength of the radiation emitted. [Take : mass of electron `= (0.5//c^(2))MeV` and `hc = 1.2 xx 10^(-12) MeV-m`, where h is the Planck's constant and c is the velocity of light in air.

Explanation

To find the wavelength of the radiation emitted when a positron and an electron collide and annihilate each other, we can follow these steps:

### Step 1: Determine the total mass-energy of the system
When a positron and an electron collide, they annihilate each other, and their total mass-energy is converted into energy. The total mass of the system can be expressed as:
\[ M = 2m_e \]
where \( m_e \) is the mass of the electron (and also the mass of the positron).

### Step 2: Substitute the mass of the electron
Given that the mass of the electron is provided as:
\[ m_e = \frac{0.5}{c^2} \text{ MeV} \]
Thus, the total mass can be calculated as:
\[ M = 2 \times \frac{0.5}{c^2} = \frac{1}{c^2} \text{ MeV} \]

### Step 3: Calculate the total energy produced
The total energy \( E \) produced from the annihilation can be calculated using the formula:
\[ E = Mc^2 \]
Substituting \( M \) into the equation gives:
\[ E = \left(\frac{1}{c^2}\right) c^2 = 1 \text{ MeV} \]

### Step 4: Determine the energy of one photon
Since the total energy is emitted in the form of two photons, the energy of one photon \( E_{\text{photon}} \) is:
\[ E_{\text{photon}} = \frac{E}{2} = \frac{1 \text{ MeV}}{2} = 0.5 \text{ MeV} \]

### Step 5: Use the energy-wavelength relationship
The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
Rearranging this gives:
\[ \lambda = \frac{hc}{E} \]

### Step 6: Substitute the values
We know:
- \( h c = 1.2 \times 10^{-12} \text{ MeV-m} \)
- \( E = 0.5 \text{ MeV} \)

Substituting these values into the equation gives:
\[ \lambda = \frac{1.2 \times 10^{-12} \text{ MeV-m}}{0.5 \text{ MeV}} = 2.4 \times 10^{-12} \text{ m} \]

### Step 7: Convert to picometers
Since \( 1 \text{ picometer} = 10^{-12} \text{ m} \), we can express the wavelength as:
\[ \lambda = 2.4 \text{ picometers} \]

### Final Answer
The wavelength of the radiation emitted is:
\[ \lambda = 2.4 \times 10^{-12} \text{ m} \quad \text{or} \quad 2.4 \text{ pm} \]

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