For pair production i.e. for the production of electron and positron, the incident photon must have a minimum frequency of the order of

Options

A. `10^(18) s^(-1)`
B. `10^(21) s^(-1)`
C. `10^(25) s^(-1)`
D. `10^(30)s^(-1)`

Answer

To solve the problem of determining the minimum frequency of a photon required for pair production of an electron and a positron, we can follow these steps:

### Step 1: Understand Pair Production
Pair production occurs when a photon interacts with a strong electromagnetic field (usually near a nucleus) to produce a particle-antiparticle pair, such as an electron and a positron. The energy of the photon must be sufficient to create these particles.

### Step 2: Calculate the Rest Mass Energy
The rest mass energy (E) of a particle is given by the equation:
\[ E = mc^2 \]
where:
- \( m \) is the mass of the particle,
- \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

For an electron and a positron:
- Mass of an electron (\( m_e \)) = \( 9.11 \times 10^{-31} \, \text{kg} \)
- Mass of a positron (\( m_p \)) = \( 9.11 \times 10^{-31} \, \text{kg} \)

The total rest mass energy for the electron-positron pair is:
\[ E_{total} = 2 \times m_e c^2 \]

### Step 3: Convert Mass to Energy in MeV
Using the conversion factor \( 1 \, \text{kg} = 6.242 \times 10^{12} \, \text{MeV} \):
\[ E_{total} = 2 \times (9.11 \times 10^{-31} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \]
Calculating this gives:
\[ E_{total} \approx 2 \times 0.511 \, \text{MeV} = 1.022 \, \text{MeV} \]

### Step 4: Relate Energy to Frequency
The energy of a photon is also related to its frequency (\( f \)) by the equation:
\[ E = hf \]
where \( h \) is Planck's constant (\( h \approx 6.626 \times 10^{-34} \, \text{Js} \)).

### Step 5: Calculate Minimum Frequency
Setting the energy equal to the photon energy, we have:
\[ hf = 1.022 \, \text{MeV} \]
To find the frequency:
\[ f = \frac{E}{h} = \frac{1.022 \times 10^6 \, \text{eV}}{6.626 \times 10^{-34} \, \text{Js}} \]

### Step 6: Convert MeV to Joules
Using the conversion \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ 1.022 \, \text{MeV} = 1.022 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J} \]

### Step 7: Substitute and Calculate
Substituting this back into the frequency equation:
\[ f = \frac{1.022 \times 10^6 \times 1.6 \times 10^{-19}}{6.626 \times 10^{-34}} \]

### Step 8: Final Calculation
Calculating this gives:
\[ f \approx 2.5 \times 10^{20} \, \text{Hz} \]

### Conclusion
The minimum frequency of the incident photon required for pair production of an electron and positron is of the order of \( 10^{20} \, \text{Hz} \).