`gamma-rays` radiation can be used to create electron-positron pair. In this process of pair production. `gamma-rays` energy cannot be less than.

Options

A. 5.0 MeV
B. 4.02 MeV
C. 15.0 MeV
D. 1.02 MeV

Explanation

To determine the minimum energy required for gamma rays to create an electron-positron pair through pair production, we can follow these steps:

### Step 1: Understand Pair Production
Pair production is a process where a photon (in this case, gamma rays) creates a particle-antiparticle pair, specifically an electron and a positron.

**Hint:** Remember that pair production requires sufficient energy to create two particles.

### Step 2: Calculate the Energy Requirement
The rest mass energy of an electron (and a positron, which has the same mass as an electron) can be calculated using the formula:
\[ E = mc^2 \]
where:
- \( m \) is the mass of the electron (approximately \( 9.11 \times 10^{-31} \) kg),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s).

The rest mass energy of one electron is:
\[ E_e = (9.11 \times 10^{-31} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2 \approx 8.19 \times 10^{-14} \text{ J} \]

Since we need to create both an electron and a positron, the total rest mass energy required is:
\[ E_{total} = 2 \times E_e \approx 2 \times 8.19 \times 10^{-14} \text{ J} \approx 1.64 \times 10^{-13} \text{ J} \]

### Step 3: Convert Energy to Mega Electron Volts (MeV)
To convert joules to electron volts (eV), we use the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Thus,
\[ E_{total} \text{ (in eV)} = \frac{1.64 \times 10^{-13} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 1.025 \times 10^6 \text{ eV} \approx 1.02 \text{ MeV} \]

### Step 4: Conclusion
The minimum energy of gamma rays required for pair production to occur is:
\[ \text{Minimum Energy} = 1.02 \text{ MeV} \]

Thus, the gamma rays' energy cannot be less than 1.02 MeV for pair production to take place.

**Final Answer:** The energy of gamma rays cannot be less than **1.02 MeV**.