Mass-Energy and Nuclear Binding Energy

Mass-Energy And Nuclear Binding Energy

Classical physics treated mass and energy as separate quantities, each conserved independently. However, Einstein's theory of special relativity revolutionized this concept, establishing the equivalence of mass and energy. This profound relationship is particularly important in nuclear physics, where mass changes are associated with enormous energy releases or absorptions, most notably in the concept of nuclear binding energy.

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Mass – Energy Equivalence ($ E=mc^2 $)

One of the most famous equations in physics is Einstein's mass-energy equivalence relation:

$ E = mc^2 $

This equation states that mass ($m$) is a form of energy, and energy ($E$) has inertia and is equivalent to mass. Here, $c$ is the speed of light in vacuum. This equation implies that a small amount of mass is equivalent to a very large amount of energy, since $c^2$ is a very large number ($c \approx 3 \times 10^8 \, m/s$, $c^2 \approx 9 \times 10^{16} \, m^2/s^2$).

Mass can be converted into energy, and energy can be converted into mass.

Examples of mass-energy conversion:

• Nuclear Reactions: In nuclear fission and fusion, a small amount of mass is converted into a large amount of kinetic energy of the resulting particles and radiation. This is the source of energy in nuclear reactors and nuclear weapons, and the energy source of stars.
• Annihilation: When a particle meets its antiparticle (e.g., an electron meets a positron), they annihilate, converting their entire mass into energy in the form of photons (gamma rays).
• Pair Production: A high-energy photon can interact with matter to create a particle-antiparticle pair (e.g., an electron and a positron), converting energy into mass.

In nuclear physics, mass differences in nuclear reactions or in the formation of nuclei are directly related to the energy released or absorbed, calculated using $E = mc^2$.

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Nuclear Binding Energy (Binding Energy per Nucleon Curve)

The nucleus of an atom is composed of protons and neutrons (nucleons) held together by the strong nuclear force. A fundamental observation in nuclear physics is that the mass of a stable nucleus is always less than the sum of the masses of its individual constituent nucleons (protons and neutrons) when they are free and separated. This difference in mass is called the mass defect ($\Delta m$).

Consider a nucleus $_Z^A X$. It contains $Z$ protons and $N = A - Z$ neutrons.

Mass of $Z$ protons = $Z \cdot m_p$

Mass of $N$ neutrons = $N \cdot m_n$

Total mass of free nucleons = $Z \cdot m_p + N \cdot m_n$

Let $M_{nucleus}$ be the actual measured mass of the nucleus $_Z^A X$. (Note: Atomic mass $M_{atom}$ usually includes the mass of electrons. $M_{nucleus} \approx M_{atom} - Z \cdot m_e$).

The mass defect ($\Delta m$) is the difference between the total mass of the individual nucleons and the actual mass of the nucleus:

$ \Delta m = (Z \cdot m_p + N \cdot m_n) - M_{nucleus} $

This mass defect represents the mass that has been converted into energy when the nucleons came together to form the nucleus. This energy is the nuclear binding energy ($E_B$).

$ E_B = \Delta m \cdot c^2 = [(Z \cdot m_p + N \cdot m_n) - M_{nucleus}] c^2 $

The binding energy is the energy required to break the nucleus into its constituent nucleons and separate them to infinite distance. It is a measure of the stability of the nucleus. A larger binding energy means the nucleons are more tightly bound, and the nucleus is more stable.

Binding energy is often expressed in units of Mega-electron Volts (MeV). $1 \, u$ is equivalent to a certain amount of energy: $1 \, u \approx 1.6605 \times 10^{-27} \, kg$. Energy equivalent of 1 u = $(1.6605 \times 10^{-27} \, kg) \times (3 \times 10^8 \, m/s)^2$ $= 1.6605 \times 10^{-27} \times 9 \times 10^{16} \, J = 14.9445 \times 10^{-11} \, J$. Convert Joules to MeV: $1 \, MeV = 1.602 \times 10^{-13} \, J$. $ \text{Energy equivalent of 1 u} = \frac{14.9445 \times 10^{-11} \, J}{1.602 \times 10^{-13} \, J/MeV} \approx 931.5 \, MeV $.

So, when calculating binding energy, if the mass defect $\Delta m$ is in atomic mass units (u), the binding energy in MeV is:

$ E_B = \Delta m \, (in \, u) \times 931.5 \, MeV/u $

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Binding Energy Per Nucleon ($ E_{B/A} $)

The total binding energy of a nucleus increases with the number of nucleons. To compare the stability of different nuclei, especially those with different mass numbers, we consider the binding energy per nucleon ($E_{B/A}$). This is the total binding energy divided by the number of nucleons ($A$):

$ E_{B/A} = \frac{E_B}{A} $

Binding energy per nucleon represents the average energy required to remove a single nucleon from the nucleus. A higher binding energy per nucleon indicates greater stability of the nucleus.

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Binding Energy per Nucleon Curve

A plot of binding energy per nucleon ($E_{B/A}$) versus mass number ($A$) provides valuable insights into nuclear stability and nuclear reactions.

Binding Energy per Nucleon curve as a function of mass number A.

Key features of the binding energy per nucleon curve:

• The curve rises sharply from light nuclei, reaches a broad maximum, and then slowly decreases for heavier nuclei.
• There are sharp peaks for very light nuclei like $^4He$, $^{12}C$, $^{16}O$, indicating their exceptionally high stability compared to their neighbours.
• The curve peaks around $A \approx 50 - 60$, specifically at $^{56}Fe$ (Iron-56), which has the highest binding energy per nucleon (about 8.79 MeV/nucleon). Nuclei around this mass number are the most stable.
• The binding energy per nucleon is approximately constant for a wide range of nuclei ($A \approx 30$ to $170$), around 8 MeV/nucleon. This indicates that the nuclear force is short-ranged; each nucleon primarily interacts with its nearest neighbours, and adding more nucleons beyond a certain number doesn't significantly increase the binding energy per nucleon (saturation of nuclear force).

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Implications for Nuclear Reactions

The shape of the binding energy curve explains why certain nuclear reactions release energy:

• Nuclear Fission: Heavy nuclei ($A > 170$) have lower binding energy per nucleon than nuclei in the peak region. If a heavy nucleus (like Uranium) splits into two or more lighter nuclei (fission fragments), the resulting lighter nuclei have higher binding energy per nucleon. The increase in binding energy per nucleon is released as energy.
• Nuclear Fusion: Very light nuclei ($A < 20$) have lower binding energy per nucleon than heavier nuclei. If two or more light nuclei combine to form a heavier nucleus (fusion), the resulting heavier nucleus has higher binding energy per nucleon (up to the peak). The increase in binding energy per nucleon is released as energy.

Thus, both fission of heavy nuclei and fusion of light nuclei are processes that move the nucleus towards the region of maximum stability (around $A=56$) and are accompanied by the release of large amounts of energy, derived from the conversion of mass defect into binding energy.

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Answer:

Given:

Helium nucleus, $_2^4He$. Atomic number $Z=2$, Mass number $A=4$. Number of protons = 2. Number of neutrons $N = A - Z = 4 - 2 = 2$.

Mass of Helium nucleus, $M_{nucleus} = 4.00260 \, u$

Mass of proton, $m_p = 1.00728 \, u$

Mass of neutron, $m_n = 1.00867 \, u$

Energy equivalent of 1 u = 931.5 MeV/c$^2$ (or 931.5 MeV if we use E=mc$^2$ directly with mass in u).

Calculate the total mass of the constituent nucleons:

Total mass of nucleons = $Z \cdot m_p + N \cdot m_n = 2 \times (1.00728 \, u) + 2 \times (1.00867 \, u)$

Total mass of nucleons = $2.01456 \, u + 2.01734 \, u = 4.03190 \, u$

Calculate the mass defect ($\Delta m$):

$ \Delta m = (\text{Total mass of nucleons}) - (\text{Mass of nucleus}) $

$ \Delta m = 4.03190 \, u - 4.00260 \, u = 0.02930 \, u $

The mass defect of the Helium nucleus is 0.02930 atomic mass units.

Calculate the binding energy ($E_B$) using the mass defect and the energy equivalent of 1 u:

$ E_B = \Delta m \times 931.5 \, MeV/u $

$ E_B = 0.02930 \, u \times 931.5 \, MeV/u $

$ E_B \approx 27.306 \, MeV $

The binding energy of the Helium nucleus is approximately 27.306 MeV.

Binding energy per nucleon = $E_B / A = 27.306 \, MeV / 4 \approx 6.826 \, MeV/nucleon$. This is a relatively high value, consistent with the high stability of the Helium nucleus (alpha particle), which shows a peak in the binding energy curve for light nuclei.