Problem 48 - Fundamentals of Physics

Question

How much energy is released when a \({ }^{238} \mathrm{U}\) nucleus decays by emitting (a) an alpha particle and (b) a sequence of neutron, proton, neutron, proton? (c) Convince yourself both by reasoned argument and by direct calculation that the difference between these two numbers is just the total binding energy of the alpha particle. (d) Find that binding energy. Some needed atomic and particle masses are $$ \begin{array}{llll} { }^{238} \mathrm{U} & 238.05079 \mathrm{u} & { }^{234} \mathrm{Th} & 234.04363 \mathrm{u} \\ { }^{237} \mathrm{U} & 237.04873 \mathrm{u} & { }^{4} \mathrm{He} & 4.00260 \mathrm{u} \\ { }^{236} \mathrm{~Pa} & 236.04891 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ { }^{235} \mathrm{~Pa} & 235.04544 \mathrm{u} & \mathrm{n} & 1.00866 \mathrm{u} \end{array} $$

Step-by-Step Solution

Answer

The energy released by the decay with an alpha particle is 4.25 MeV, while the neutron-proton sequence releases 37.98 MeV, with the binding energy of the alpha particle being 33.73 MeV.

Step 1: Identify Reaction Products for Alpha Decay

For part (a), the decay of uclear{^{238}U} by emitting an alpha particle uclear{^{4}He} results in a uclear{^{234}Th} nucleus. The reaction is:\[ ^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \]The energy released is the mass defect converted to energy.

Step 2: Calculate Mass Defect for Alpha Decay

The mass defect \(\Delta m\) is calculated by subtracting the sum of the products' masses from the original mass:\[ \Delta m = (m_{^{238}U} - (m_{^{234}Th} + m_{^{4}He})) \]Substitute the given masses:\[ \Delta m = 238.05079\,\text{u} - (234.04363\,\text{u} + 4.00260\,\text{u}) \]\[ \Delta m = 0.00456\,\text{u} \]

Step 3: Convert Mass Defect to Energy for Alpha Decay

Use the energy-mass equivalence \(E = \Delta m \cdot c^2\), where \(c = 931.5\,\text{MeV/u}\):\[ E = 0.00456\,\text{u} \times 931.5\,\text{MeV/u} = 4.25\,\text{MeV} \]

Step 4: Identify Reaction Products for Neutron-Proton Decay Sequence

For part (b), follow the decay:1. \(^{238}U_{92} \rightarrow ^{237}U_{92} + n\) (neutron emission)2. \(^{237}U_{92} \rightarrow ^{236}Pa_{91} + p\) (proton emission)3. \(^{236}Pa_{91} \rightarrow ^{235}Pa_{91} + n\)4. \(^{235}Pa_{91} \rightarrow ^{234}Th_{90} + p\)Thus, the final reaction is:\[ ^{238}U_{92} \rightarrow ^{234}Th_{90} + 2n + 2p \]

Step 5: Calculate Mass Defect for Neutron-Proton Decay Sequence

The total mass defect \(\Delta m\) is:\[ \Delta m = (m_{^{238}U} - (m_{^{234}Th} + 2m_n + 2m_p)) \]Substitute the given masses:\[ \Delta m = 238.05079\,\text{u} - (234.04363\,\text{u} + 2 \times 1.00866\,\text{u} + 2 \times 1.00783\,\text{u}) \]\[ \Delta m = 0.04078\,\text{u} \]

Step 6: Convert Mass Defect to Energy for Neutron-Proton Sequence

Use the energy-mass equivalence:\[ E = 0.04078\,\text{u} \times 931.5\,\text{MeV/u} = 37.98\,\text{MeV} \]

Step 7: Difference in Energies and Binding Energy of Alpha Particle

For part (c) and (d), the difference between the energy released in both processes is:\[ \Delta E = 37.98\,\text{MeV} - 4.25\,\text{MeV} = 33.73\,\text{MeV} \]This difference is the binding energy of the alpha particle. Thus, the binding energy of the alpha particle is 28.73 MeV.

Key Concepts

Alpha Particle

An alpha particle consists of two protons and two neutrons, making it identical to a helium nucleus ( uc{4}{2}{He} ). It is a type of ionizing radiation, commonly emitted during the decay of heavy elements like uranium isotopes. Alpha particles are relatively large and heavy compared to other types of radiation, such as beta particles or gamma rays.
Their size and charge mean they have low penetration power, being able to be stopped by a sheet of paper or the outer layer of human skin. However, if ingested or inhaled, they can be extremely harmful due to their high ionizing capability.
When a uranium-238 nucleus ( uc{238}{92}{U} ) emits an alpha particle, it transforms into thorium-234 ( uc{234}{90}{Th} ), reflecting the loss of two protons and two neutrons. This process helps stabilize heavy nuclei by reducing the nuclear size. Understanding the emission of alpha particles is crucial in fields such as nuclear physics and energy, especially concerning nuclear decay and energy release.

Binding Energy

Binding energy is the energy required to disassemble a nucleus into its separate protons and neutrons, essentially opposing the attractive forces that hold the nucleus together. It is also the energy released when a nucleus is formed from these particles. The higher the binding energy, the more stable the nucleus.
For the alpha particle, this binding energy reveals its stability, acting as a mini nucleus within larger atoms. The binding energy of an alpha particle is about 28.73 MeV, showing how tightly bound its protons and neutrons are.
In nuclear decay, comparing the binding energies before and after a decay event helps determine the energy released, providing insights into both the process's efficiency and the nucleus's stability. Harnessing this knowledge is essential when considering reactions like nuclear fission and fusion, where energy changes play a key role.

Mass Defect

Mass defect refers to the difference in mass between a nucleus and its component nucleons (protons and neutrons). This discrepancy arises because some mass is converted into binding energy, as described by Einstein's mass-energy equivalence formula, \[E = \Delta m \cdot c^2\] where \(c\) is the speed of light.
The mass defect becomes apparent when calculating the mass of a uranium-238 nucleus compared with the mass of its decay products. For example, in the alpha decay of uranium-238, the mass defect is the difference between the uranium nucleus's mass and the combined masses of the thorium nucleus and the alpha particle.
This mass defect translates into the energy released during the decay, further demonstrating how nuclear reactions can release enormous amounts of energy. In practical terms, it's a key concept in nuclear physics, important for understanding energy output in nuclear reactors as well as naturally occurring radioactive decay processes.

Uranium-238 Decay

Uranium-238 is a radioactive isotope that undergoes decay to reach a more stable state. It decays primarily through alpha emission, where it loses two protons and two neutrons to form thorium-234. This natural process contributes to uranium's long half-life of about 4.5 billion years, making it a prevalent element in the Earth's crust.
The decay sequence can continue as thorium-234 itself is radioactive and further decays, eventually leading down a chain of transformations known as the uranium series or decay chain, culminating in stable lead-206. This chain involves various decay modes, such as beta decay and further alpha decays, illustrating the complex nature of nuclear stability.
Understanding uranium-238's decay is crucial not only for managing radioactive materials but also for applications in radiometric dating, nuclear power, and more. By tracing these decay processes, scientists can date geological formations and archaeological finds or harness the energy released in nuclear power generation.