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Atoms and Nuclei - Mass-Energy Relation and Binding Energy

Grade 12ICSEPhysics

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

🔑Concepts

Einstein's Mass-Energy Equivalence: Mass and energy are inter-convertible. A loss in mass appears as energy and vice versa, governed by the relation E=mc2E = mc^2.

Atomic Mass Unit (amuamu or uu): It is defined as 112\frac{1}{12}th of the mass of one atom of carbon-12 (612C^{12}_{6}C). 1 u1.660539×1027 kg1\ u \approx 1.660539 \times 10^{-27}\ kg.

Energy Equivalent of 1 u1\ u: The energy released by the total conversion of 1 u1\ u of mass into energy is approximately 931.5 MeV931.5\ 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 always positive for a stable nucleus.

Binding Energy (BEBE): The energy required to break a nucleus into its constituent nucleons, or the energy released when nucleons combine to form a nucleus. It is the energy equivalent of the mass defect.

Binding Energy per Nucleon (BE\overline{BE}): Defined as BEA\frac{BE}{A}, where AA is the mass number. It is a measure of the stability of the nucleus. A higher value indicates a more stable nucleus.

Binding Energy Curve: A plot of BEA\frac{BE}{A} against mass number AA. It shows that nuclei with 30<A<17030 < A < 170 are most stable, with 56Fe^{56}Fe being one of the most stable. Very light nuclei (A<30A < 30) can undergo fusion, and very heavy nuclei (A>170A > 170) can undergo fission to increase stability.

📐Formulae

E=mc2E = mc^2

1 u=1.6605×1027 kg1\ u = 1.6605 \times 10^{-27}\ kg

1 u931.5 MeV1\ u \approx 931.5\ MeV

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

BE=Δm×c2BE = \Delta m \times c^2

BE(in MeV)=Δm(in u)×931.5 MeVBE (in\ MeV) = \Delta m (in\ u) \times 931.5\ MeV

BE=BEA\overline{BE} = \frac{BE}{A}

💡Examples

Problem 1:

Calculate the binding energy and binding energy per nucleon of an alpha particle (24He^4_2He). Given: mass of proton mp=1.007276 um_p = 1.007276\ u, mass of neutron mn=1.008665 um_n = 1.008665\ u, and mass of helium nucleus M=4.001506 uM = 4.001506\ u.

Solution:

  1. Identify constituents: Z=2Z = 2 protons and (AZ)=2(A - Z) = 2 neutrons.
  2. Calculate Mass Defect (Δm\Delta m): Δm=[2×mp+2×mn]M\Delta m = [2 \times m_p + 2 \times m_n] - M Δm=[2×1.007276+2×1.008665]4.001506\Delta m = [2 \times 1.007276 + 2 \times 1.008665] - 4.001506 Δm=[2.014552+2.017330]4.001506=0.030376 u\Delta m = [2.014552 + 2.017330] - 4.001506 = 0.030376\ u
  3. Calculate Total Binding Energy (BEBE): BE=Δm×931.5 MeV=0.030376×931.5=28.295 MeVBE = \Delta m \times 931.5\ MeV = 0.030376 \times 931.5 = 28.295\ MeV
  4. Calculate Binding Energy per Nucleon (BE\overline{BE}): BE=BEA=28.2954=7.07 MeV/nucleon\overline{BE} = \frac{BE}{A} = \frac{28.295}{4} = 7.07\ MeV/nucleon.

Explanation:

The mass defect represents the 'missing' mass converted into energy to hold the nucleons together. Dividing this energy by the total number of nucleons (4 for Helium) gives the stability per nucleon.

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