Sure! Let's break down the process of calculating the energy released in the fission process step-by-step.
### Step 1: Determine the Mass Defect
The mass defect is the difference in mass between the initial Plutonium nucleus and the total mass of the fragments. This mass defect is the mass converted into energy during the fission process.
Given:
- Mass of Plutonium nucleus, [tex]\( m_{\text{plutonium}} = 4.986 \times 10^{-27} \)[/tex] kg
- Total mass of fragments, [tex]\( m_{\text{fragments}} = 4.198 \times 10^{-27} \)[/tex] kg
The mass defect, [tex]\( \Delta m \)[/tex], is calculated as:
[tex]\[ \Delta m = m_{\text{plutonium}} - m_{\text{fragments}} \][/tex]
Substituting in the given values:
[tex]\[ \Delta m = 4.986 \times 10^{-27} \, \text{kg} - 4.198 \times 10^{-27} \, \text{kg} \][/tex]
[tex]\[ \Delta m = 7.88 \times 10^{-28} \, \text{kg} \][/tex]
### Step 2: Calculate the Energy Released in Joules
According to Einstein's mass-energy equivalence principle (E=mc²), the energy released, [tex]\( E \)[/tex], can be calculated using the mass defect and the speed of light.
Given:
- Speed of light, [tex]\( c = 3 \times 10^8 \)[/tex] m/s
The energy released, [tex]\( E \)[/tex], is given by:
[tex]\[ E = \Delta m \cdot c^2 \][/tex]
Substituting the values we have:
[tex]\[ E = 7.88 \times 10^{-28} \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]
[tex]\[ E = 7.88 \times 10^{-28} \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ E = 7.092 \times 10^{-11} \, \text{J} \][/tex]
So, the energy released in Joules is:
[tex]\[ E \approx 7.09 \times 10^{-11} \, \text{J} \][/tex]
### Step 3: Convert the Energy to MeV
To convert the energy from Joules to MeV, we use the conversion factor:
[tex]\[ 1 \, \text{J} \approx 6.242 \times 10^{12} \, \text{MeV} \][/tex]
The energy in MeV, [tex]\( E_{\text{MeV}} \)[/tex], is:
[tex]\[ E_{\text{MeV}} = E \times 6.242 \times 10^{12} \][/tex]
Substituting the value of [tex]\( E \)[/tex] from earlier:
[tex]\[ E_{\text{MeV}} = 7.09 \times 10^{-11} \, \text{J} \times 6.242 \times 10^{12} \, \text{MeV/J} \][/tex]
[tex]\[ E_{\text{MeV}} \approx 442.68 \, \text{MeV} \][/tex]
So, the energy released in MeV is:
[tex]\[ E \approx 442.68 \, \text{MeV} \][/tex]
### Summary
(a) The energy released in the fission process is approximately [tex]\( 7.09 \times 10^{-11} \)[/tex] Joules.
(b) The energy released in the fission process is approximately [tex]\( 442.68 \)[/tex] MeV.
