Answer :
Sure, let's go through each part of the question step-by-step.
### a) Write the equation of the nuclear fission reaction that took place. (2 marks)
The question states that Uranium-235 (U-235) undergoes fission and produces Barium-141 (Ba-141), Krypton-92 (Kr-92), and 2 neutrons. The equation of the nuclear fission reaction is:
[tex]\[ \text{U-235} \rightarrow \text{Ba-141} + \text{Kr-92} + 2 \text{neutrons} \][/tex]
### b) Determine the mass defect. (2 marks)
The mass defect ([tex]\(\Delta m\)[/tex]) is the difference between the mass of the reactants and the mass of the products. Here are the given masses in kg:
- Mass of Uranium-235, [tex]\( m_{\text{U-235}} = 390.2480 \times 10^{-27} \)[/tex] kg
- Mass of Barium-141, [tex]\( m_{\text{Ba-141}} = 233.9616 \times 10^{-27} \)[/tex] kg
- Mass of Krypton-92, [tex]\( m_{\text{Kr-92}} = 152.5794 \times 10^{-27} \)[/tex] kg
- Mass of a neutron, [tex]\( m_{\text{neutron}} = 1.6747 \times 10^{-27} \)[/tex] kg
The initial mass, which is the mass of U-235:
[tex]\[ m_{\text{initial}} = m_{\text{U-235}} = 390.2480 \times 10^{-27} \text{ kg} \][/tex]
The final mass, which includes the masses of Ba-141, Kr-92, and 2 neutrons:
[tex]\[ m_{\text{final}} = m_{\text{Ba-141}} + m_{\text{Kr-92}} + 2 \times m_{\text{neutron}} \][/tex]
[tex]\[ m_{\text{final}} = 233.9616 \times 10^{-27} \text{ kg} + 152.5794 \times 10^{-27} \text{ kg} + 2 \times 1.6747 \times 10^{-27} \text{ kg} \][/tex]
[tex]\[ m_{\text{final}} = 233.9616 \times 10^{-27} \text{ kg} + 152.5794 \times 10^{-27} \text{ kg} + 3.3494 \times 10^{-27} \text{ kg} \][/tex]
[tex]\[ m_{\text{final}} = 389.8904 \times 10^{-27} \text{ kg} \][/tex]
The mass defect is:
[tex]\[ \Delta m = m_{\text{initial}} - m_{\text{final}} \][/tex]
[tex]\[ \Delta m = 390.2480 \times 10^{-27} \text{ kg} - 389.8904 \times 10^{-27} \text{ kg} \][/tex]
[tex]\[ \Delta m = 3.575999999999832 \times 10^{-28} \text{ kg} \][/tex]
### c) Calculate the energy released by the reduction in mass in the reaction in MeV. (3 marks)
The energy released can be calculated using Einstein's mass-energy equivalence principle [tex]\( E = \Delta m \, c^2 \)[/tex], where [tex]\( c \)[/tex] is the speed of light ( [tex]\( c = 3 \times 10^8 \text{ m/s} \)[/tex]). Additionally, we need to convert the energy from joules to MeV. The conversion factor is [tex]\( 1 \text{ kg m}^2/\text{s}^2 = 5.60958884 \times 10^{26} \text{ MeV} \)[/tex].
First, calculate the energy in joules:
[tex]\[ E = \Delta m \, c^2 \][/tex]
[tex]\[ E = 3.575999999999832 \times 10^{-28} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \][/tex]
[tex]\[ E = 3.575999999999832 \times 10^{-28} \times 9 \times 10^{16} \][/tex]
[tex]\[ E = 3.218399999999849 \times 10^{-11} \text{ J} \][/tex]
Convert joules to MeV:
[tex]\[ E (\text{MeV}) = 3.218399999999849 \times 10^{-11} \text{ J} \times 5.60958884 \times 10^{26} \text{ MeV/J} \][/tex]
[tex]\[ E (\text{MeV}) = 1.805390072265515 \times 10^{16} \text{ MeV} \][/tex]
Thus, the energy released in the fission reaction is [tex]\( 1.805390072265515 \times 10^{16} \text{ MeV} \)[/tex].
### a) Write the equation of the nuclear fission reaction that took place. (2 marks)
The question states that Uranium-235 (U-235) undergoes fission and produces Barium-141 (Ba-141), Krypton-92 (Kr-92), and 2 neutrons. The equation of the nuclear fission reaction is:
[tex]\[ \text{U-235} \rightarrow \text{Ba-141} + \text{Kr-92} + 2 \text{neutrons} \][/tex]
### b) Determine the mass defect. (2 marks)
The mass defect ([tex]\(\Delta m\)[/tex]) is the difference between the mass of the reactants and the mass of the products. Here are the given masses in kg:
- Mass of Uranium-235, [tex]\( m_{\text{U-235}} = 390.2480 \times 10^{-27} \)[/tex] kg
- Mass of Barium-141, [tex]\( m_{\text{Ba-141}} = 233.9616 \times 10^{-27} \)[/tex] kg
- Mass of Krypton-92, [tex]\( m_{\text{Kr-92}} = 152.5794 \times 10^{-27} \)[/tex] kg
- Mass of a neutron, [tex]\( m_{\text{neutron}} = 1.6747 \times 10^{-27} \)[/tex] kg
The initial mass, which is the mass of U-235:
[tex]\[ m_{\text{initial}} = m_{\text{U-235}} = 390.2480 \times 10^{-27} \text{ kg} \][/tex]
The final mass, which includes the masses of Ba-141, Kr-92, and 2 neutrons:
[tex]\[ m_{\text{final}} = m_{\text{Ba-141}} + m_{\text{Kr-92}} + 2 \times m_{\text{neutron}} \][/tex]
[tex]\[ m_{\text{final}} = 233.9616 \times 10^{-27} \text{ kg} + 152.5794 \times 10^{-27} \text{ kg} + 2 \times 1.6747 \times 10^{-27} \text{ kg} \][/tex]
[tex]\[ m_{\text{final}} = 233.9616 \times 10^{-27} \text{ kg} + 152.5794 \times 10^{-27} \text{ kg} + 3.3494 \times 10^{-27} \text{ kg} \][/tex]
[tex]\[ m_{\text{final}} = 389.8904 \times 10^{-27} \text{ kg} \][/tex]
The mass defect is:
[tex]\[ \Delta m = m_{\text{initial}} - m_{\text{final}} \][/tex]
[tex]\[ \Delta m = 390.2480 \times 10^{-27} \text{ kg} - 389.8904 \times 10^{-27} \text{ kg} \][/tex]
[tex]\[ \Delta m = 3.575999999999832 \times 10^{-28} \text{ kg} \][/tex]
### c) Calculate the energy released by the reduction in mass in the reaction in MeV. (3 marks)
The energy released can be calculated using Einstein's mass-energy equivalence principle [tex]\( E = \Delta m \, c^2 \)[/tex], where [tex]\( c \)[/tex] is the speed of light ( [tex]\( c = 3 \times 10^8 \text{ m/s} \)[/tex]). Additionally, we need to convert the energy from joules to MeV. The conversion factor is [tex]\( 1 \text{ kg m}^2/\text{s}^2 = 5.60958884 \times 10^{26} \text{ MeV} \)[/tex].
First, calculate the energy in joules:
[tex]\[ E = \Delta m \, c^2 \][/tex]
[tex]\[ E = 3.575999999999832 \times 10^{-28} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \][/tex]
[tex]\[ E = 3.575999999999832 \times 10^{-28} \times 9 \times 10^{16} \][/tex]
[tex]\[ E = 3.218399999999849 \times 10^{-11} \text{ J} \][/tex]
Convert joules to MeV:
[tex]\[ E (\text{MeV}) = 3.218399999999849 \times 10^{-11} \text{ J} \times 5.60958884 \times 10^{26} \text{ MeV/J} \][/tex]
[tex]\[ E (\text{MeV}) = 1.805390072265515 \times 10^{16} \text{ MeV} \][/tex]
Thus, the energy released in the fission reaction is [tex]\( 1.805390072265515 \times 10^{16} \text{ MeV} \)[/tex].
