What is the approximate energy released by the complete annihilation of an alpha particle?

Options

A. 931 J
B. 372 J
C. 931 MeV
D. 3724 MeV

Answer

To find the approximate energy released by the complete annihilation of an alpha particle, we can follow these steps:

### Step 1: Understand the Composition of an Alpha Particle
An alpha particle consists of 2 protons and 2 neutrons. This gives it a total atomic mass of approximately 4 atomic mass units (amu).

### Step 2: Use Einstein's Mass-Energy Equivalence
According to Einstein's equation, the energy (E) released when a mass (m) is annihilated is given by:
\[ E = mc^2 \]
where \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s).

### Step 3: Energy Released per Atomic Mass Unit
It is known that the energy released when 1 amu is converted to energy is approximately 931.5 MeV (mega electron volts).

### Step 4: Calculate the Total Energy for the Alpha Particle
Since the alpha particle has a mass of 4 amu, the total energy released by the annihilation of the alpha particle can be calculated as:
\[ E = 4 \times 931.5 \text{ MeV} \]
Calculating this gives:
\[ E = 3726 \text{ MeV} \]

### Step 5: Convert Energy to Joules (if necessary)
To convert MeV to Joules, we use the conversion factor:
\[ 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ Joules} \]
Thus, the energy in Joules is:
\[ E = 3726 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} \]
Calculating this gives:
\[ E \approx 5.9616 \times 10^{-10} \text{ Joules} \]

### Final Answer
The approximate energy released by the complete annihilation of an alpha particle is:
- **3726 MeV** or approximately **\( 5.96 \times 10^{-10} \) Joules**.

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