If a proton and anti-proton come close to each other and annihilate, how much energy will be released ?

Options

A. `1.5 xx 10^10 J`
B. `3 xx 10^-10 J`
C. `4.5 xx 10^-10 J`
D. None of these

Explanation

To find the energy released when a proton and an anti-proton annihilate, we can follow these steps:

### Step 1: Determine the mass of the proton and anti-proton
The mass of a proton (m_p) is approximately \(1.67 \times 10^{-27} \, \text{kg}\). Since the mass of the anti-proton is equal to the mass of the proton, we have:
\[ m_{\text{anti-proton}} = m_p = 1.67 \times 10^{-27} \, \text{kg} \]

### Step 2: Calculate the total mass
When a proton and an anti-proton annihilate, their total mass (m_total) is the sum of their individual masses:
\[ m_{\text{total}} = m_p + m_{\text{anti-proton}} = 1.67 \times 10^{-27} \, \text{kg} + 1.67 \times 10^{-27} \, \text{kg} = 2 \times 1.67 \times 10^{-27} \, \text{kg} = 3.34 \times 10^{-27} \, \text{kg} \]

### Step 3: Convert mass to energy using Einstein's equation
According to Einstein's mass-energy equivalence principle, the energy (E) released can be calculated using the formula:
\[ E = mc^2 \]
where \(c\) is the speed of light, approximately \(3 \times 10^8 \, \text{m/s}\).

### Step 4: Calculate the energy
Substituting the total mass into the equation:
\[ E = (3.34 \times 10^{-27} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \]
Calculating \(c^2\):
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]
Now substituting back:
\[ E = 3.34 \times 10^{-27} \times 9 \times 10^{16} = 3.006 \times 10^{-10} \, \text{J} \]

### Step 5: Convert energy to mega electron volts
To convert joules to mega electron volts (MeV), we use the conversion factor \(1 \, \text{J} = 6.242 \times 10^{12} \, \text{MeV}\):
\[ E = 3.006 \times 10^{-10} \, \text{J} \times 6.242 \times 10^{12} \, \text{MeV/J} \approx 1.88 \times 10^{3} \, \text{MeV} \approx 1862 \, \text{MeV} \]

### Conclusion
The energy released when a proton and an anti-proton annihilate is approximately \(1862 \, \text{MeV}\) or \(3 \times 10^{-10} \, \text{J}\).