A positron of 1MeV collides with an electron of 1MeV and gets annihilated and the reaction produces two `gamma-`ray photons. If the effective mass of each photons is 0.0016amu, then the energy of each `gamma-`ray photon is about-

Options

A. 1.5 MeV
B. 3MeV
C. 6MeV
D. 2 MeV

Answer

To solve the problem step by step, we will calculate the total energy involved in the annihilation of the positron and electron, and then determine the energy of each gamma-ray photon produced.

### Step 1: Determine the total energy of the positron
The total energy of the positron can be calculated by adding its kinetic energy and its rest mass energy.

- Kinetic energy of the positron = 1 MeV
- Rest mass energy of the positron = 0.512 MeV

Total energy of the positron:
\[ E_{positron} = KE_{positron} + E_{rest\ mass} = 1\ MeV + 0.512\ MeV = 1.512\ MeV \]

### Step 2: Determine the total energy of the electron
Similarly, we can calculate the total energy of the electron.

- Kinetic energy of the electron = 1 MeV
- Rest mass energy of the electron = 0.512 MeV

Total energy of the electron:
\[ E_{electron} = KE_{electron} + E_{rest\ mass} = 1\ MeV + 0.512\ MeV = 1.512\ MeV \]

### Step 3: Calculate the total energy before annihilation
The total energy before annihilation is the sum of the total energies of the positron and the electron.

\[ E_{total} = E_{positron} + E_{electron} = 1.512\ MeV + 1.512\ MeV = 3.024\ MeV \]

### Step 4: Determine the energy of each gamma-ray photon
Since the annihilation produces two identical gamma-ray photons, the energy of each photon can be found by dividing the total energy by 2.

\[ E_{photon} = \frac{E_{total}}{2} = \frac{3.024\ MeV}{2} = 1.512\ MeV \]

### Step 5: Approximate the energy of each gamma-ray photon
Rounding off the energy to a reasonable number of significant figures, we get:

\[ E_{photon} \approx 1.5\ MeV \]

### Final Answer
The energy of each gamma-ray photon is approximately **1.5 MeV**.

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