Answer :
### (a) Nuclear Reaction
The nuclear reaction involves an alpha particle (Helium-4 nucleus) interacting with a Sodium-23 nucleus to produce a Silicon-27 nucleus and energy. The reaction can be written as:
[tex]\[ \text{Na-23} + \text{He-4} \rightarrow \text{Si-27} + Q \][/tex]
where [tex]\( Q \)[/tex] represents the energy released during the reaction.
### (b) Calculating the Energy Released
1. Determine the Masses of Reactants and Products:
- Mass of Sodium-23 (Na) = 23.1004 u
- Mass of Helium-4 (He) = 4.0783 u
- Mass of Silicon-27 (Si) = 26.886 u
2. Calculate the Mass Defect:
The mass defect ([tex]\( \Delta m \)[/tex]) is the difference between the total mass of the reactants and the mass of the product:
[tex]\[ \Delta m = ( \text{Mass of Sodium} + \text{Mass of Helium} ) - \text{Mass of Silicon} \][/tex]
Substituting the values:
[tex]\[ \Delta m = (23.1004 \, \text{u} + 4.0783 \, \text{u}) - 26.886 \, \text{u} = 0.2927 \, \text{u} \][/tex]
3. Convert the Mass Defect to Kilograms:
Given [tex]\(1 \, \text{u} = 1.67 \times 10^{-27} \, \text{kg}\)[/tex],
[tex]\[ \Delta m_{\text{kg}} = 0.2927 \, \text{u} \times 1.67 \times 10^{-27} \, \text{kg/u} = 4.88809 \times 10^{-28} \, \text{kg} \][/tex]
4. Calculate the Energy Released per Sodium Atom:
Using the formula [tex]\( E = \Delta m c^2 \)[/tex] where [tex]\( c = 3.0 \times 10^8 \, \text{m/s} \)[/tex],
[tex]\[ E_{\text{per atom}} = 4.88809 \times 10^{-28} \, \text{kg} \times (3.0 \times 10^8 \, \text{m/s})^2 = 4.39928 \times 10^{-11} \, \text{Joules} \][/tex]
5. Calculate the Number of Silicon-27 Atoms in Given Mass:
Given the mass of Silicon-27 = 26.886 u, convert to kilograms:
[tex]\[ \text{Mass of one Silicon-27 atom} = 26.886 \, \text{u} \times 1.67 \times 10^{-27} \, \text{kg/u} = 4.48900 \times 10^{-26} \, \text{kg} \][/tex]
Number of Silicon-27 atoms in [tex]\(2.7 \times 10^{-15} \, \text{kg}\)[/tex]:
[tex]\[ \text{Number of atoms} = \frac{2.7 \times 10^{-15} \, \text{kg}}{4.48900 \times 10^{-26} \, \text{kg/atom}} = 6.01341 \times 10^{10} \, \text{atoms} \][/tex]
6. Calculate the Total Energy Released:
Total energy is the product of the energy per atom and the number of atoms:
[tex]\[ E_{\text{total}} = 4.39928 \times 10^{-11} \, \text{J/atom} \times 6.01341 \times 10^{10} \, \text{atoms} = 2.64547 \, \text{J} \][/tex]
### Final Answer
The energy released by [tex]\(2.7 \times 10^{-15} \, \text{kg}\)[/tex] of Silicon-27 is approximately [tex]\(2.645 \, \text{J}\)[/tex].
The nuclear reaction involves an alpha particle (Helium-4 nucleus) interacting with a Sodium-23 nucleus to produce a Silicon-27 nucleus and energy. The reaction can be written as:
[tex]\[ \text{Na-23} + \text{He-4} \rightarrow \text{Si-27} + Q \][/tex]
where [tex]\( Q \)[/tex] represents the energy released during the reaction.
### (b) Calculating the Energy Released
1. Determine the Masses of Reactants and Products:
- Mass of Sodium-23 (Na) = 23.1004 u
- Mass of Helium-4 (He) = 4.0783 u
- Mass of Silicon-27 (Si) = 26.886 u
2. Calculate the Mass Defect:
The mass defect ([tex]\( \Delta m \)[/tex]) is the difference between the total mass of the reactants and the mass of the product:
[tex]\[ \Delta m = ( \text{Mass of Sodium} + \text{Mass of Helium} ) - \text{Mass of Silicon} \][/tex]
Substituting the values:
[tex]\[ \Delta m = (23.1004 \, \text{u} + 4.0783 \, \text{u}) - 26.886 \, \text{u} = 0.2927 \, \text{u} \][/tex]
3. Convert the Mass Defect to Kilograms:
Given [tex]\(1 \, \text{u} = 1.67 \times 10^{-27} \, \text{kg}\)[/tex],
[tex]\[ \Delta m_{\text{kg}} = 0.2927 \, \text{u} \times 1.67 \times 10^{-27} \, \text{kg/u} = 4.88809 \times 10^{-28} \, \text{kg} \][/tex]
4. Calculate the Energy Released per Sodium Atom:
Using the formula [tex]\( E = \Delta m c^2 \)[/tex] where [tex]\( c = 3.0 \times 10^8 \, \text{m/s} \)[/tex],
[tex]\[ E_{\text{per atom}} = 4.88809 \times 10^{-28} \, \text{kg} \times (3.0 \times 10^8 \, \text{m/s})^2 = 4.39928 \times 10^{-11} \, \text{Joules} \][/tex]
5. Calculate the Number of Silicon-27 Atoms in Given Mass:
Given the mass of Silicon-27 = 26.886 u, convert to kilograms:
[tex]\[ \text{Mass of one Silicon-27 atom} = 26.886 \, \text{u} \times 1.67 \times 10^{-27} \, \text{kg/u} = 4.48900 \times 10^{-26} \, \text{kg} \][/tex]
Number of Silicon-27 atoms in [tex]\(2.7 \times 10^{-15} \, \text{kg}\)[/tex]:
[tex]\[ \text{Number of atoms} = \frac{2.7 \times 10^{-15} \, \text{kg}}{4.48900 \times 10^{-26} \, \text{kg/atom}} = 6.01341 \times 10^{10} \, \text{atoms} \][/tex]
6. Calculate the Total Energy Released:
Total energy is the product of the energy per atom and the number of atoms:
[tex]\[ E_{\text{total}} = 4.39928 \times 10^{-11} \, \text{J/atom} \times 6.01341 \times 10^{10} \, \text{atoms} = 2.64547 \, \text{J} \][/tex]
### Final Answer
The energy released by [tex]\(2.7 \times 10^{-15} \, \text{kg}\)[/tex] of Silicon-27 is approximately [tex]\(2.645 \, \text{J}\)[/tex].
