How much energy is released when 1 amu of mass is annihilated ?

Options

A. `931.5 eV`
B. `1.49 xx 10 ^(-3)J`
C. `14.138 xx 10 ^(-17)kWh`
D. all of these

Answer

To calculate the energy released when 1 atomic mass unit (amu) of mass is annihilated, we can use Einstein's mass-energy equivalence principle, which is given by the equation:

\[ E = mc^2 \]

Where:
- \( E \) is the energy,
- \( m \) is the mass (in kilograms),
- \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \) m/s).

### Step 1: Convert 1 amu to kilograms
1 atomic mass unit (amu) is defined as:

\[ 1 \text{ amu} = 1.660539 \times 10^{-27} \text{ kg} \]

### Step 2: Use the mass-energy equivalence formula
Now, we will substitute the mass in kilograms into the equation \( E = mc^2 \):

\[ E = (1.660539 \times 10^{-27} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2 \]

### Step 3: Calculate \( c^2 \)
First, calculate \( c^2 \):

\[ c^2 = (3 \times 10^8 \text{ m/s})^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]

### Step 4: Substitute \( c^2 \) back into the energy equation
Now substitute \( c^2 \) back into the energy equation:

\[ E = (1.660539 \times 10^{-27} \text{ kg}) \times (9 \times 10^{16} \text{ m}^2/\text{s}^2) \]

### Step 5: Perform the multiplication
Now, we can perform the multiplication:

\[ E = 1.660539 \times 9 \times 10^{-11} \text{ kg m}^2/\text{s}^2 \]

Calculating \( 1.660539 \times 9 \):

\[ E = 14.944851 \times 10^{-11} \text{ J} \]

### Step 6: Convert joules to electron volts
To convert joules to electron volts, we use the conversion factor:

\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]

Thus,

\[ E = \frac{14.944851 \times 10^{-11} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \]

### Step 7: Calculate the final energy in eV
Calculating the above expression gives:

\[ E \approx 933 \times 10^6 \text{ eV} \]

### Final Result
The energy released when 1 amu of mass is annihilated is approximately:

\[ E \approx 933 \text{ MeV} \]