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Understand the concept of Binding Energy.

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2 parts
i

5.7071 MeV/nucleon\text{5.7071 MeV/nucleon}

ii

7.0747 MeV/nucleon\text{7.0747 MeV/nucleon}

Step-by-step solution

10 steps

Step 1: Understand the concept of Binding Energy.
The binding energy (BE) of a nucleus is the energy required to separate it into its constituent protons and neutrons. It is calculated from the mass defect (Δm\Delta m), which is the difference between the total mass of the individual nucleons and the actual mass of the nucleus.
The relationship is given by Einstein's mass-energy equivalence: BE=Δmc2BE = \Delta m c^2.
In nuclear physics, it's common to use the conversion factor 1 amu=931.5 MeV1 \text{ amu} = 931.5 \text{ MeV}.
The binding energy per nucleon is the total binding energy divided by the number of nucleons (protons + neutrons) in the nucleus.

Given masses:
Mass of deuteron M(12H)=2.013554 amuM(^2_1H) = 2.013554 \text{ amu}
Mass of alpha particle M(24He)=4.002604 amuM(^4_2He) = 4.002604 \text{ amu}
Mass of proton Mp=1.007825 amuM_p = 1.007825 \text{ amu}
Mass of neutron Mn=1.008665 amuM_n = 1.008665 \text{ amu}

Part i· 5 steps

Calculate the binding energy per nucleon when two deuterons combine to form an α\alpha-particle.

The nuclear reaction is: 12H+12H→24He^2_1H + ^2_1H \to ^4_2He

  1. Calculate the total mass of the reactants.

    Mreactants=2×M(12H)M_{\text{reactants}} = 2 \times M(^2_1H) Mreactants=2×2.013554 amuM_{\text{reactants}} = 2 \times 2.013554 \text{ amu} Mreactants=4.027108 amuM_{\text{reactants}} = 4.027108 \text{ amu}

  2. Identify the mass of the product.

    Mproduct=M(24He)=4.002604 amuM_{\text{product}} = M(^4_2He) = 4.002604 \text{ amu}

  3. Calculate the mass defect (Δm\Delta m).

    Δm=Mreactants−Mproduct\Delta m = M_{\text{reactants}} - M_{\text{product}} Δm=4.027108 amu−4.002604 amu\Delta m = 4.027108 \text{ amu} - 4.002604 \text{ amu} Δm=0.024504 amu\Delta m = 0.024504 \text{ amu}

  4. Calculate the total binding energy (BE).

    BE=Δm×931.5MeVamuBE = \Delta m \times 931.5 \frac{\text{MeV}}{\text{amu}} BE=0.024504 amu×931.5MeVamuBE = 0.024504 \text{ amu} \times 931.5 \frac{\text{MeV}}{\text{amu}} BE=22.828276 MeVBE = 22.828276 \text{ MeV}

  5. Calculate the binding energy per nucleon.

    The α\alpha-particle (24He^4_2He) has 4 nucleons (2 protons and 2 neutrons).

    BEper nucleon=BENumber of nucleonsBE_{\text{per nucleon}} = \frac{BE}{\text{Number of nucleons}} BEper nucleon=22.828276 MeV4BE_{\text{per nucleon}} = \frac{22.828276 \text{ MeV}}{4} 5.7071 MeV/nucleon\boxed{\text{5.7071 MeV/nucleon}}

Part ii

7.0747 MeV/nucleon\text{7.0747 MeV/nucleon}

Calculate the binding energy per nucleon when two neutrons and two protons combine to form an α\alpha-particle.

The nuclear reaction is: 2n+2p→24He2n + 2p \to ^4_2He

  1. Calculate the total mass of the reactants.

    Mreactants=(2×Mn)+(2×Mp)M_{\text{reactants}} = (2 \times M_n) + (2 \times M_p) Mreactants=(2×1.008665 amu)+(2×1.007825 amu)M_{\text{reactants}} = (2 \times 1.008665 \text{ amu}) + (2 \times 1.007825 \text{ amu}) Mreactants=2.017330 amu+2.015650 amuM_{\text{reactants}} = 2.017330 \text{ amu} + 2.015650 \text{ amu} Mreactants=4.032980 amuM_{\text{reactants}} = 4.032980 \text{ amu}

  2. Identify the mass of the product.

    Mproduct=M(24He)=4.002604 amuM_{\text{product}} = M(^4_2He) = 4.002604 \text{ amu}

  3. Calculate the mass defect (Δm\Delta m).

    Δm=Mreactants−Mproduct\Delta m = M_{\text{reactants}} - M_{\text{product}} Δm=4.032980 amu−4.002604 amu\Delta m = 4.032980 \text{ amu} - 4.002604 \text{ amu} Δm=0.030376 amu\Delta m = 0.030376 \text{ amu}

  4. Calculate the total binding energy (BE).

    BE=Δm×931.5MeVamuBE = \Delta m \times 931.5 \frac{\text{MeV}}{\text{amu}} BE=0.030376 amu×931.5MeVamuBE = 0.030376 \text{ amu} \times 931.5 \frac{\text{MeV}}{\text{amu}} BE=28.298964 MeVBE = 28.298964 \text{ MeV}

  5. Calculate the binding energy per nucleon.

    The α\alpha-particle (24He^4_2He) has 4 nucleons.

    BEper nucleon=BENumber of nucleonsBE_{\text{per nucleon}} = \frac{BE}{\text{Number of nucleons}} BEper nucleon=28.298964 MeV4BE_{\text{per nucleon}} = \frac{28.298964 \text{ MeV}}{4} 7.0747 MeV/nucleon\boxed{\text{7.0747 MeV/nucleon}}
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