Chap.20 Fission and fusion//Ex.

  

Exercises

Chap.20

💓Exercise 1:

Alpha and Gamma Decay of Bismuth-212

Bismuth-212 (²¹²₈₃Bi) undergoes both alpha and gamma decay. The daughter nucleus is a thallium (Tl) isotope.

Given Data:

• Mass of ²¹²₈₃Bi = 211.991876 u
• Mass of Tl = 207.982013 u
• Mass of α particle = 4.0015 u
• 1 u = 931.5 MeV/c²
• Wavelength of γ (λ_γ) = 3.78 pm = 3.78×10^-12 m
• Planck's constant (h) = 6.63×10^-34 J·s
• Speed of light (c) = 3×10⁸ m/s

Questions:

1. What causes the gamma radiation emission?
2. Write the gamma de-excitation equation (1)
3. Write the alpha decay equation (2) of bismuth
4. Calculate the energy released in nuclear reaction (2)
5. Determine the kinetic energy and velocity of the alpha particle

💓Exercise 2: Energy Release in Fusion

Calculate the energy released when two deuterium nuclei fuse to form helium-4:

2 ²H → ⁴He

Given the binding energies:

• BE(²H) = 2.23 MeV
• BE(⁴He) = 28.30 MeV

Solution:

Energy released = BE(products) - BE(reactants)

💓 Exercise 3:       Binding Energy and Stability of Nuclei

Consider the isotopes Helium-3 (\(^{3}_{2}He\)) and Helium-4 (\(^{4}_{2}He\)). You are given the following atomic masses:

• Mass of Helium-3 (\(m(^{3}_{2}He)\)): 3.01603 u
• Mass of Helium-4 (\(m(^{4}_{2}He)\)): 4.00260 u
• Mass of a proton (\(m_p\)): 1.00728 u
• Mass of a neutron (\(m_n\)): 1.00867 u
• 1 u = 931.5 MeV/c²

Your tasks are to:

1. Calculate the mass defect (\(\Delta m\)) for both Helium-3 and Helium-4 nuclei.
2. Calculate the total binding energy (\(E_b\)) (in MeV) for both Helium-3 and Helium-4 nuclei.
3. Calculate the binding energy per nucleon (\(E_b/A\)) (in MeV/nucleon) for both Helium-3 and Helium-4 nuclei.
4. Based on your results in part (c), which of the two isotopes is more stable? Explain your reasoning.

💓Exercise 4:      Evolution of the Sun's Mass During Its Lifetime

The sun, with a current mass of \(M_s = 2 \times 10^{30}\) kg, is a giant nuclear fusion reactor that produces a radiative power of \(P = 3.9 \times 10^{26}\) W, assumed constant.

1. Calculate the mass lost by the sun given that its age is \(4.5 \times 10^{9}\) years.
2. One of the nuclear fusion reactions produced in the sun's core, transforming hydrogen into helium, is: \[ 4 \ ^1_1H \longrightarrow \ ^4_2He + 2 \ ^0_{+1}e + 2 \nu + 25.7 \text{ MeV} \] The mass of the core (of which 35% by mass is hydrogen, 64% by mass is helium, and 1% by mass is other elements) represents one-eighth of the sun's mass.
   1. Calculate the number of helium nuclei produced by the total transformation of hydrogen.
   2. Deduce the radiant energy produced by the transformation of hydrogen into helium.
3. Hydrogen runs out; the sun is formed of helium by fusion, and helium transforms into carbon according to the reaction: \[ 3 \ ^4_2He \longrightarrow \ ^{12}_6C + \gamma + 7.27 \text{ MeV} \]
   1. Calculate the total number of helium nuclei in the core.
   2. Deduce the radiant energy from the total transformation of helium into carbon.
4. The sun dies when its core contains only carbon.
   1. Calculate the radiant energy from the sun from now until its death.
   2. After how much time will the sun die?

Solution: Evolution of the Sun's Mass During Its Lifetime

1. Calculation of the mass lost by the sun:

   The total energy radiated by the sun during its age is: \[ E_{total} = P \times t = (3.9 \times 10^{26} \text{ W}) \times (4.5 \times 10^{9} \text{ years} \times 3.154 \times 10^{7} \text{ s/year}) \] \[ E_{total} \approx 5.53 \times 10^{43} \text{ J} \] Using the mass-energy equivalence \(E = mc^2\), the mass lost is: \[ \Delta m = \frac{E_{total}}{c^2} = \frac{5.53 \times 10^{43} \text{ J}}{(3 \times 10^{8} \text{ m/s})^2} \approx 6.14 \times 10^{26} \text{ kg} \]

2. Transformation of hydrogen into helium:
   1. Number of helium nuclei produced:

      Mass of hydrogen in the core: \(m_H = 0.35 \times \frac{1}{8} \times M_s = 0.35 \times \frac{1}{8} \times 2 \times 10^{30} \text{ kg} = 8.75 \times 10^{28} \text{ kg}\)

      Number of moles of hydrogen: \(n_H = \frac{m_H}{M_{mol}(H)} = \frac{8.75 \times 10^{28} \text{ kg}}{1.008 \times 10^{-3} \text{ kg/mol}} \approx 8.68 \times 10^{31} \text{ mol}\)

      Number of hydrogen atoms: \(N_H = n_H \times N_A = (8.68 \times 10^{31} \text{ mol}) \times (6.022 \times 10^{23} \text{ mol}^{-1}) \approx 5.23 \times 10^{55}\)

      For every 4 hydrogen nuclei, 1 helium nucleus is produced. So, the number of helium nuclei produced is: \[ N_{He} = \frac{N_H}{4} = \frac{5.23 \times 10^{55}}{4} \approx 1.31 \times 10^{55} \text{ nuclei} \]

   2. Radiant energy produced:

      The energy released per helium nucleus formed is 25.7 MeV. The total radiant energy is: \[ E_{radiated\_H} = N_{He} \times 25.7 \text{ MeV} = (1.31 \times 10^{55}) \times (25.7 \times 1.602 \times 10^{-13} \text{ J/MeV}) \approx 5.40 \times 10^{43} \text{ J} \]

3. Transformation of helium into carbon:
   1. Total number of helium nuclei in the core:

      Initial mass of helium in the core: \(m_{He\_initial} = 0.64 \times \frac{1}{8} \times M_s = 0.64 \times \frac{1}{8} \times 2 \times 10^{30} \text{ kg} = 1.6 \times 10^{29} \text{ kg}\)

      Mass of helium formed from hydrogen fusion: \(m_{He\_formed} = N_{He} \times M_{nucleus}(He) = (1.31 \times 10^{55}) \times (4.0026 \text{ u} \times 1.66 \times 10^{-27} \text{ kg/u}) \approx 8.70 \times 10^{28} \text{ kg}\)

      Total mass of helium in the core before carbon transformation: \(m_{He\_total} = m_{He\_initial} + m_{He\_formed} = 1.6 \times 10^{29} \text{ kg} + 8.70 \times 10^{28} \text{ kg} = 2.47 \times 10^{29} \text{ kg}\)

      Total number of helium nuclei: \(N_{He\_total} = \frac{m_{He\_total}}{M_{nucleus}(He)} = \frac{2.47 \times 10^{29} \text{ kg}}{4.0026 \times 1.66 \times 10^{-27} \text{ kg}} \approx 3.73 \times 10^{55} \text{ nuclei}\)

   2. Radiant energy from the total transformation of helium into carbon:

      For every 3 helium nuclei, the energy released is 7.27 MeV. The number of reactions is \(N_{C\_formed} = \frac{N_{He\_total}}{3} \approx 1.24 \times 10^{55}\)

      The total radiant energy is: \[ E_{radiated\_He} = N_{C\_formed} \times 7.27 \text{ MeV} = (1.24 \times 10^{55}) \times (7.27 \times 1.602 \times 10^{-13} \text{ J/MeV}) \approx 1.44 \times 10^{43} \text{ J} \]

4. Death of the sun (carbon core):
   1. Total radiant energy from the sun until its death:

      The total radiant energy will be approximately the sum of the energy from hydrogen and helium fusion: \[ E_{total\_life} = E_{radiated\_H} + E_{radiated\_He} = 5.40 \times 10^{43} \text{ J} + 1.44 \times 10^{43} \text{ J} \approx 6.84 \times 10^{43} \text{ J} \]

   2. Time before the sun dies:

      Assuming a constant radiative power, the total time before the sun dies is: \[ t_{death} = \frac{E_{total\_life}}{P} = \frac{6.84 \times 10^{43} \text{ J}}{3.9 \times 10^{26} \text{ W}} \approx 1.75 \times 10^{17} \text{ s} \] \[ t_{death} \approx \frac{1.75 \times 10^{17}}{3.154 \times 10^{7} \text{ s/year}} \approx 5.55 \times 10^{9} \text{ years} \]

⬆️ العودة إلى الأعلى