Assume that a neutron breaks into a proton and an electron . The energy reased during this process is (mass of neutron `= 1.6725 xx 10^(-27) kg` mass of proton `= 1.6725 xx 10^(-27) kg` mass of electron `= 9 xx 10^(-31) kg )`

Options

A. `0.73 MeV`
B. `7.10 MeV`
C. `6.30 MeV`
D. `5.4 MeV`

Explanation

To find the energy released when a neutron breaks into a proton and an electron, we can follow these steps:

### Step 1: Identify the masses involved
We have the following masses:
- Mass of neutron, \( m_n = 1.6725 \times 10^{-27} \) kg
- Mass of proton, \( m_p = 1.6725 \times 10^{-27} \) kg
- Mass of electron, \( m_e = 9 \times 10^{-31} \) kg

### Step 2: Calculate the mass defect
The mass defect (\( \Delta m \)) is calculated as follows:
\[ \Delta m = m_n - (m_p + m_e) \]
Substituting the values:
\[ \Delta m = 1.6725 \times 10^{-27} - (1.6725 \times 10^{-27} + 9 \times 10^{-31}) \]
\[ \Delta m = 1.6725 \times 10^{-27} - 1.6725 \times 10^{-27} - 9 \times 10^{-31} \]
\[ \Delta m = -9 \times 10^{-31} \text{ kg} \]
Since the mass of the proton and neutron are equal, the mass defect simplifies to:
\[ \Delta m = 9 \times 10^{-31} \text{ kg} \]

### Step 3: Calculate the energy released using Einstein's equation
The energy released (\( E \)) can be calculated using the formula:
\[ E = \Delta m c^2 \]
Where \( c \) is the speed of light, approximately \( 3 \times 10^8 \) m/s. Substituting the values:
\[ E = (9 \times 10^{-31}) \times (3 \times 10^8)^2 \]
Calculating \( (3 \times 10^8)^2 \):
\[ (3 \times 10^8)^2 = 9 \times 10^{16} \]
Now substituting back:
\[ E = 9 \times 10^{-31} \times 9 \times 10^{16} \]
\[ E = 81 \times 10^{-15} \text{ Joules} \]

### Step 4: Convert energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules} \):
\[ E_{eV} = \frac{81 \times 10^{-15}}{1.6 \times 10^{-19}} \]
Calculating this gives:
\[ E_{eV} = \frac{81}{1.6} \times 10^{4} \approx 50.625 \times 10^{4} \text{ eV} \]
\[ E_{eV} \approx 5.0625 \times 10^{5} \text{ eV} \]

### Step 5: Convert to Mega electron volts
To convert to Mega electron volts (MeV), divide by \( 10^6 \):
\[ E_{MeV} = \frac{5.0625 \times 10^{5}}{10^6} \approx 0.50625 \text{ MeV} \]

### Final Answer
The energy released during the process is approximately \( 0.506 \text{ MeV} \).

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