A `5 xx 10^(-4) Å` photon produces an electron-positron pair in the vincinity of a heavy nucleus. Rest energy of electron is `0.511 MeV`. sIf they have the same kinetic energies, the energy of each paricles is nearly

Options

A. `1.2 MeV`
B. `12 MeV`
C. `120MeV`
D. `1200MeV`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Energy of the Photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the photon (\(5 \times 10^{-4} \, \text{m}\)).

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{5 \times 10^{-4} \, \text{m}} \]
Calculating this gives:
\[ E \approx 3.976 \times 10^{-14} \, \text{J} \]

### Step 2: Convert Energy from Joules to MeV
To convert the energy from Joules to MeV, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\) and \(1 \, \text{MeV} = 10^6 \, \text{eV}\):
\[ E \approx \frac{3.976 \times 10^{-14} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 248.5 \, \text{MeV} \]

### Step 3: Calculate the Total Energy of the Electron-Positron Pair
The rest energy of an electron (and positron) is given as \(0.511 \, \text{MeV}\). For a pair, the total rest energy is:
\[ E_{\text{rest}} = 2 \times 0.511 \, \text{MeV} = 1.022 \, \text{MeV} \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the total energy of the photon is equal to the sum of the rest energy and the kinetic energy of the particles:
\[ E_{\text{photon}} = E_{\text{rest}} + 2K \]
where \(K\) is the kinetic energy of each particle. Rearranging gives:
\[ K = \frac{E_{\text{photon}} - E_{\text{rest}}}{2} \]

### Step 5: Substitute Values and Solve for K
Substituting the values:
\[ K = \frac{248.5 \, \text{MeV} - 1.022 \, \text{MeV}}{2} = \frac{247.478 \, \text{MeV}}{2} \approx 123.739 \, \text{MeV} \]

### Step 6: Final Answer
Thus, the energy of each particle (electron and positron) is approximately:
\[ K \approx 123.739 \, \text{MeV} \]