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Nuclei - Mass-Energy Relation and Binding Energy

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Einstein's Mass-Energy Equivalence: Mass is a form of energy and can be converted into other forms of energy. The relation is given by E=mc2E = mc^2.

Atomic Mass Unit (uu): Defined as 112\frac{1}{12}th of the mass of a carbon-12 (612C{}^{12}_{6}C) atom. 1 u=1.660539×1027 kg1 \text{ u} = 1.660539 \times 10^{-27} \text{ kg}.

Mass-Energy Equivalent of 1 u1 \text{ u}: The energy equivalent of one atomic mass unit is approximately 931.5 MeV931.5 \text{ MeV}.

Mass Defect (Δm\Delta m): The difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual rest mass of the nucleus. It is expressed as Δm=[Zmp+(AZ)mn]M\Delta m = [Z m_p + (A - Z) m_n] - M.

Binding Energy (EbE_b): The energy required to break a nucleus into its constituent nucleons, or the energy released when nucleons combine to form a nucleus. It is calculated as Eb=Δmc2E_b = \Delta m c^2.

Binding Energy per Nucleon (EbnE_{bn}): It is the average energy required to extract one nucleon from the nucleus. Ebn=EbAE_{bn} = \frac{E_b}{A}. It is a direct measure of nuclear stability.

Stability and EbnE_{bn} Curve: Nuclei with higher EbnE_{bn} are more stable. The curve peaks around A56A \approx 56 (Iron, 56Fe{}^{56}Fe), where Ebn8.8 MeVE_{bn} \approx 8.8 \text{ MeV}.

Nuclear Fission and Fusion: Very heavy nuclei (A>170A > 170) undergo fission to increase EbnE_{bn}, while very light nuclei (A<10A < 10) undergo fusion to increase EbnE_{bn}, both processes releasing large amounts of energy.

📐Formulae

E=mc2E = mc^2

Δm=[Zmp+(AZ)mn]Mnucleus\Delta m = [Z m_p + (A - Z) m_n] - M_{nucleus}

Eb=Δm×931.5 MeV(if Δm is in u)E_b = \Delta m \times 931.5 \text{ MeV} \quad \text{(if } \Delta m \text{ is in u)}

Ebn=EbAE_{bn} = \frac{E_b}{A}

1 eV=1.602×1019 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}

💡Examples

Problem 1:

Calculate the binding energy and binding energy per nucleon of an alpha particle (24He{}^{4}_{2}He). Given: mass of proton mp=1.007276 um_p = 1.007276 \text{ u}, mass of neutron mn=1.008665 um_n = 1.008665 \text{ u}, and mass of helium nucleus M=4.001506 uM = 4.001506 \text{ u}.

Solution:

  1. Calculate Mass Defect (Δm\Delta m): Δm=[Zmp+(AZ)mn]M\Delta m = [Z m_p + (A - Z) m_n] - M Δm=[2(1.007276)+2(1.008665)]4.001506\Delta m = [2(1.007276) + 2(1.008665)] - 4.001506 Δm=[2.014552+2.017330]4.001506\Delta m = [2.014552 + 2.017330] - 4.001506 Δm=4.0318824.001506=0.030376 u\Delta m = 4.031882 - 4.001506 = 0.030376 \text{ u}

  2. Calculate Binding Energy (EbE_b): Eb=Δm×931.5 MeVE_b = \Delta m \times 931.5 \text{ MeV} Eb=0.030376×931.528.3 MeVE_b = 0.030376 \times 931.5 \approx 28.3 \text{ MeV}

  3. Calculate Binding Energy per Nucleon (EbnE_{bn}): Ebn=EbA=28.34=7.075 MeV/nucleonE_{bn} = \frac{E_b}{A} = \frac{28.3}{4} = 7.075 \text{ MeV/nucleon}

Explanation:

First, we find the difference between the theoretical mass of constituents and the actual mass. This missing mass is converted into binding energy according to Einstein's relation. Dividing by the number of nucleons gives the stability per nucleon.

Mass-Energy Relation and Binding Energy Revision - Class 12 Physics CBSE