AP Physics 2- 15.7 Fission, Fusion, and Nuclear Decay- Study Notes- New Syllabus
AP Physics 2- 15.7 Fission, Fusion, and Nuclear Decay – Study Notes
AP Physics 2- 15.7 Fission, Fusion, and Nuclear Decay – Study Notes – per latest Syllabus.
Key Concepts:
- Nuclear Reactions and the Strong Force
- Nuclear Fission and Fusion
- Radioactive Decay, Half-Life, and Decay Constant
Nuclear Reactions and the Strong Force
Nuclear reactions involve interactions between nuclei or nucleons, governed by the strong nuclear force and constrained by fundamental conservation laws. Energy can be released during these reactions due to mass-energy equivalence.
The Strong Nuclear Force
The strong force acts at very short ranges (nuclear scale, ~1 fm). It binds protons and neutrons (nucleons) together in the nucleus, overcoming electrostatic repulsion between protons also dominates nucleon interactions within the nucleus.
Conservation Laws in Nuclear Reactions
- Nucleon number: $\text{Total number of protons + neutrons remains constant. }$
- Energy: Total energy is conserved; kinetic energy and photon emission are included.
- Momentum: Total momentum of the system is conserved.
- Mass-energy equivalence: Mass can be converted to energy and vice versa: $\mathrm{E = m c^2} $
- Energy released may appear as kinetic energy of products or as photons (gamma rays).
Note:
- The strong force confines nucleons and is the main force responsible for nuclear stability.
- Conservation laws (nucleon number, energy, momentum) determine possible nuclear reactions.
- Mass defects in nuclear reactions lead to energy release according to \(\mathrm{E = mc^2}\).
- Energy can be observed as kinetic energy of products or gamma radiation.
Example :
A deuteron (\(\mathrm{^2_1H}\)) collides with a proton (\(\mathrm{^1_1H}\)) to form helium-3 (\(\mathrm{^3_2He}\)) and a gamma photon. Verify conservation of nucleon number.
▶️ Answer/Explanation
Step 1: Reaction: \(\mathrm{^2_1H + ^1_1H \to ^3_2He + \gamma}\)
Step 2: Nucleon number check: Left: 2 + 1 = 3, Right: 3 (He-3) → conserved
Step 3: Charge check: Left: 1 + 1 = 2, Right: 2 (He-3) → conserved
Example :
In alpha decay of \(\mathrm{^{238}_{92}U \to ^{234}_{90}Th + ^4_2He}\), calculate the energy released if the mass defect is \(0.004 \, u\).
▶️ Answer/Explanation
Step 1: Convert mass defect to kg: \(\mathrm{0.004 \, u \cdot 1.6605 \times 10^{-27} \, kg/u \approx 6.642 \times 10^{-30} \, kg}\)
Step 2: Energy released: \(\mathrm{E = m c^2 = 6.642 \times 10^{-30} \cdot (3 \times 10^8)^2 \approx 5.98 \times 10^{-13} \, J}\)
Step 3: Interpretation: This energy appears as kinetic energy of the alpha particle and recoiling thorium nucleus.
Nuclear Fission and Fusion
Nuclear reactions can release enormous amounts of energy through either splitting a heavy nucleus (fission) or combining light nuclei (fusion). Both processes involve the strong nuclear force and mass-energy equivalence (\(\mathrm{E=mc^2}\)).
Nuclear Fission
Fission is the splitting of a heavy nucleus (e.g., uranium-235 or plutonium-239) into two lighter nuclei, along with neutrons and energy.
General reaction:
$ \mathrm{^A_Z X + n \to ^{A_1}_{Z_1} Y + ^{A_2}_{Z_2} Z + k \, n + \text{energy}} $
Energy released comes from the mass defect between reactants and products:
\(\mathrm{E = \Delta m \, c^2}\)
- Released neutrons can induce a chain reaction in a fissile material.
- Used in nuclear reactors and atomic bombs.
- Products are radioactive and can have significant environmental impact.
Nuclear Fusion
Fusion is the combining of two light nuclei (e.g., isotopes of hydrogen) to form a heavier nucleus, with the release of energy.
General reaction (hydrogen fusion):
$ \mathrm{^2_1H + ^3_1H \to ^4_2He + n + \text{energy}} $
Energy is released due to the mass of the fused nucleus being less than the sum of its parts
(\(\mathrm{E=mc^2}\)).
- Fusion powers stars, including the Sun.
- Requires extremely high temperatures and pressures to overcome electrostatic repulsion between nuclei.
- Produces less long-lived radioactive waste compared to fission.
Key Features
- Both fission and fusion convert a small amount of mass into a large amount of energy (\(\mathrm{E = mc^2}\)).
- Fission: splits heavy nuclei, releases neutrons → chain reactions possible.
- Fusion: combines light nuclei, requires high temperature and pressure, powers stars.
- Both involve the strong nuclear force binding nucleons in the nucleus.
- Fission produces radioactive products; fusion generally produces stable helium nuclei.
Example :
Write the fission reaction for \(\mathrm{^{235}_{92}U}\) when it absorbs a neutron and produces \(\mathrm{^{141}_{56}Ba}\) and \(\mathrm{^{92}_{36}Kr}\) along with two neutrons.
▶️ Answer/Explanation
Step 1: Reaction: \(\mathrm{^{235}_{92}U + n \to ^{141}_{56}Ba + ^{92}_{36}Kr + 2 n + \text{energy}}\)
Step 2: Interpretation: Mass defect converts to kinetic energy of products and released neutrons.
Example :
Consider hydrogen fusion: \(\mathrm{^2_1H + ^3_1H \to ^4_2He + n + energy}\). Determine which particle carries most of the released energy.
▶️ Answer/Explanation
Step 1: Products: helium-4 nucleus and neutron.
Step 2: Neutron is lighter and receives most kinetic energy (due to momentum conservation).
Step 3: Helium nucleus carries remaining energy as kinetic energy.
Radioactive Decay, Half-Life, and Decay Constant
Radioactive decay is the spontaneous transformation of an unstable nucleus into one or more different nuclei, often emitting particles or photons. The decay process is random for individual nuclei but predictable statistically for large samples.
Decay Law
The number of nuclei remaining in a sample after time \(t\) is given by the exponential decay law:
\[ \mathrm{N = N_0 e^{-\lambda t}} \]
where:
- \(\mathrm{N_0}\) = initial number of nuclei,
- \(\mathrm{N}\) = number of nuclei remaining at time \(t\),
- \(\mathrm{\lambda}\) = decay constant.
Decay Constant (\(\boldsymbol{\lambda}\))
\(\lambda\) represents the probability per unit time that a nucleus will decay.
Can be used to predict the number of nuclei remaining after a certain period or estimate the age of a material if the initial quantity is known.
Related to half-life:
\[ \mathrm{t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}} \]
Half-Life (\(\boldsymbol{t_{1/2}}\))
Half-life is the time required for half the nuclei in a sample to decay. It is independent of the initial number of nuclei. Different isotopes can have vastly different half-lives, from fractions of a second to billions of years.
Derived from the decay law:
\[ \mathrm{N = N_0 \, e^{-\lambda t} \implies t_{1/2} = \frac{\ln 2}{\lambda}} \]
Important Point:
- Decay constant allows calculation of remaining nuclei or material age.
- Half-life is characteristic of each radioactive isotope.
- Exponential decay means the same fraction (50%) of nuclei decays in each half-life interval.
- Applications include radiometric dating, nuclear medicine, and reactor physics.
Example :
A sample contains 800 radioactive nuclei with decay constant \(\lambda = 0.02 \, h^{-1}\). How many nuclei remain after 50 hours?
▶️ Answer/Explanation
Step 1: Use decay law: \(\mathrm{N = N_0 e^{-\lambda t} = 800 \cdot e^{-0.02 \cdot 50}}\)
Step 2: Calculate: \(\mathrm{N \approx 800 \cdot e^{-1} \approx 800 \cdot 0.3679 \approx 294}\)
Example :
A rock contains 25% of its original carbon-14. Given the half-life of carbon-14 is 5730 years, estimate the age of the rock.
▶️ Answer/Explanation
Step 1: Use fraction remaining: \(\mathrm{N/N_0 = (1/2)^{t/t_{1/2}}}\)
Step 2: Solve for \(t\): \(\mathrm{0.25 = (1/2)^{t/5730} \implies t/5730 = 2 \implies t \approx 11460 \, years}\)