Problem 22 - Fundamentals of Physics

Question

Go An \(\alpha\) particle ('He nucleus) is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an \(\alpha\) particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass. $$ \begin{array}{llll} { }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array} $$

Step-by-Step Solution

Answer

Energy to remove a proton: ~19.8 MeV, a neutron: ~20.6 MeV, separate p-n: ~2.2 MeV. Total binding energy: ~28.3 MeV, per nucleon: ~7.1 MeV. No exact match with (a), (b), or (c).

Step 1: Introduction to Atomic Mass

Begin by understanding that the atomic mass unit (u) is a standard unit of mass that quantifies mass on an atomic or molecular scale. Elements are made of nucleons (protons and neutrons), and different isotopes have different numbers of these nucleons.

Step 1: Energy to Remove a Proton

To remove a proton from the 4\text{He}, compare the mass of helium to the sum of \{}^{3}T and {}^{1}H. Compute the energy by using \\(E_1 = (M({}^{3}\text{H}) + M({}^{1}\text{H}) - M({}^{4}\text{He}))\cdot 931.5 \, \text{MeV/u}\) where 931.5 MeV/u is the conversion factor from atomic units to energy. Substitute in the given masses. \\(E_1 = (3.01605 \, \text{u} + 1.00783 \, \text{u} - 4.00260 \, \text{u}) \cdot 931.5 \\approx 19.81 \, \text{MeV}\). The energy required to remove a proton from the \(\alpha\) particle is approximately 19.81 MeV.

Step 2: Energy to Remove a Neutron

To remove a neutron, calculate the mass difference between a helium nucleon and a deuterium nucleus next with a neutron, \(E_2 = (M({}^{2}\text{H}) + M(\text{n}) - M({}^{4}\text{He})) \cdot 931.5 \, \text{MeV/u}\). Plug in the numbers: \\(E_2 = (2.01410 \, \text{u} + 1.00867 \, \text{u} - 4.00260 \, \text{u}) \cdot 931.5 \), \resulting in \\(E_2 \approx 20.58 \, \text{MeV}\). \The energy to remove a neutron is approximately 20.58 MeV.

Step 3: Energy to Separate Proton and Neutron

To separate the last proton and neutron, compute the deuterium mass with components, \(E_3 = (M({}^{1}\text{H}) + M(\text{n}) - M({}^{2}\text{H})) \cdot 931.5 \, \text{MeV/u}\). Insert values: \\(E_3 = (1.00783 \, \text{u} + 1.00867 \, \text{u} - 2.01410 \, \text{u}) \cdot 931.5 \), \which gives \\(E_3 \approx 2.23 \, \text{MeV}\). \To separate a proton-neutron pair requires approximately 2.23 MeV.

Step 4: Total Binding Energy

Determine the total binding energy required to free all nucleons in \(B_ ext{total} = E_1 + E_2 + E_3 = 19.81 \, \text{MeV} + 20.58 \, \text{MeV} + 2.23 \, \text{MeV} \approx 42.62 \, \text{MeV}\). The total energy that binds these nucleons in an \(\alpha\) particle is approximately 28.30 MeV.

Step 5: Binding Energy Per Nucleon

For binding energy per nucleon, calculate \\(B_ ext{nucleon} = \frac{B_{total}}{4}\) because there are 4 nucleons: \\(B_ ext{nucleon} = \frac{42.62 \, \text{MeV}}{4} \approx 7.15 \, \text{MeV/nuclon}\). \The binding energy per nucleon is approximately 7.15 MeV per nucleon.

Step 6: Match Questions (a), (b), or (c)

Compare each step's energy requirement: (a) 19.81 MeV, (b) 20.58 MeV, and (c) 2.23 MeV, with the total and per-nucleon energies. None match the entire binding energy (42.62 MeV), but 'c' has a value close to one nucleon binding at 7.15 MeV per nucleon.

Key Concepts

Alpha Particle

An alpha particle is a type of nuclear particle consisting of two protons and two neutrons, closely resembling a helium nucleus. This makes it quite stable and less reactive compared to other nuclear entities. Alpha particles are commonly emitted in the form of alpha radiation during the radioactive decay of heavier elements. This process is part of how certain isotopes naturally convert over time into more stable atoms. Alpha particles play a crucial role in nuclear reactions and are often used in experiments to probe the structure of atomic nuclei.

Understanding the behavior of alpha particles helps in studying the forces that hold nuclei together and the concept of nuclear binding energy. When dealing with nuclear equations, it's important to consider the mass defect, which reflects the difference between the sum of the masses of the separate nucleons and the actual mass of the nucleus. This

Atomic Mass Unit

An atomic mass unit (amu or u) is a standard unit of mass that quantifies mass on an atomic or molecular scale. It is specifically one twelfth of the mass of an unbound carbon-12 atom in its nuclear and electronic ground state. Atomic mass units are commonly used because they allow for more manageable and relatable measurements of atomic-scale masses.

Using amu allows scientists to easily compare the masses of protons, neutrons, and entire atoms, which typically have masses in the range of single or few units. Converting these measurements to energy, as seen in nuclear reactions, utilizes the conversion factor 931.5 MeV/u. This conversion is critical in calculating nuclear binding energy, which gives insight into the stability of different isotopes and is essential for understanding reactions such as fission and fusion.

Proton and Neutron Separation

Proton and neutron separation involves removing nucleons from a nucleus, which requires a specific amount of energy known as separation energy. This process gives a direct view into the amount of binding energy that holds a nucleus together.

When separating a proton or a neutron from an alpha particle, you are effectively overcoming the nuclear forces that bind these particles together. The energy required can be calculated using the differences in atomic masses before and after the separation, converted into energy using the relationship between mass and energy.

For example, removing a proton or a neutron from a stable nucleus like an alpha particle often requires significant energy, indicating the strength of the nuclear force within the particle. This separation energy is pivotal for understanding nuclear reactions and for calculating nuclear binding energy, which can explain different behaviors and stability profiles of isotopes.

Isotopes

Isotopes are versions of elements that have the same number of protons but different numbers of neutrons. This variation in neutrons leads to differences in their nuclear masses and stability. The concept of isotopes is crucial in nuclear physics and chemistry because it helps explain why certain nuclei are more stable or more prone to decay.

Isotopes have the same chemical properties due to having the same electronic configuration but different physical properties, such as mass and nuclear stability. In nuclear exercises, isotopes are often calculated and considered to understand how the binding energy varies depending on the nucleus's composition.

Each isotope has a particular binding energy, which can be calculated using atomic masses and the conversion of mass difference into energy. This knowledge is essential when predicting the behavior of elements in different nuclear reactions, where the release or absorption of energy depends heavily on the isotopic composition.