Nuclear Physics Essentials

Cheatsheet Content

Mass Defect & Nuclear Binding Energy Energy & Mass Equivalence Einstein's theory of relativity states that mass can be converted into energy and vice-versa. This is known as mass-energy equivalence, summarized by the equation: $$E = mc^2$$ $E$ = energy (J) $m$ = mass (kg) $c$ = speed of light ($3.00 \times 10^8 \text{ m s}^{-1}$) Examples: Fusion of hydrogen into helium in the sun. Fission of uranium in nuclear power plants. Nuclear weapons. High-energy particle collisions in accelerators. Nuclear Equations Nuclear reactions are represented by balanced equations using AZX notation: $^{A}_{Z}X$ $A$ = Nucleon number (total protons + neutrons) $Z$ = Proton number (atomic number) In nuclear reactions, both nucleon number ($A$) and proton number ($Z$) must be conserved. Worked Example: If a neutron is captured by Uranium-235: $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{95}_{42}Mo + ^{139}_{57}La + x^{1}_{0}n + 7^{0}_{-1}e$ Balance nucleon numbers: $235 + 1 = 95 + 139 + x(1) + 7(0) \implies 236 = 234 + x \implies x = 2$ Mass Defect & Binding Energy Mass Defect ($ \Delta m $): The difference between an atom's mass and the sum of the masses of its constituent protons and neutrons. $$ \Delta m = Zm_p + (A-Z)m_n - M_{total} $$ $m_p$ = mass of a proton $m_n$ = mass of a neutron $M_{total}$ = measured mass of the nucleus A system of separated nucleons has a greater mass than when bound in a nucleus. This mass difference is converted to energy upon nucleus formation. Binding Energy: The energy required to break a nucleus into its constituent protons and neutrons. $$E = \Delta m c^2$$ Binding Energy per Nucleon Definition: The binding energy of a nucleus divided by the number of nucleons in the nucleus. Higher binding energy per nucleon indicates greater nuclear stability. Iron-56 ($^{56}_{26}Fe$) has the highest binding energy per nucleon, making it the most stable element. Binding Energy Curve: At low nucleon numbers ($A$): Nuclei are less stable, attractive nuclear forces dominate. Lighter elements tend to undergo fusion. At high nucleon numbers ($A$): Repulsive electrostatic forces dominate, making nuclei less stable. Heavier elements tend to undergo fission. Worked Example: Calculate the binding energy per nucleon of Iron-56 ($^{56}_{26}Fe$) in MeV. Given: $m_n = 1.675 \times 10^{-27} \text{ kg}$, $m_p = 1.673 \times 10^{-27} \text{ kg}$, $M_{Fe} = 9.288 \times 10^{-26} \text{ kg}$ Protons ($Z$) = 26, Neutrons ($A-Z$) = $56-26 = 30$ Mass defect: $ \Delta m = (26 \times 1.673 \times 10^{-27}) + (30 \times 1.675 \times 10^{-27}) - (9.288 \times 10^{-26}) = 8.680 \times 10^{-28} \text{ kg} $ Binding energy ($E = \Delta m c^2$): $E = (8.680 \times 10^{-28}) \times (3.00 \times 10^8)^2 = 7.812 \times 10^{-11} \text{ J}$ Binding energy per nucleon: $E/A = (7.812 \times 10^{-11}) / 56 = 1.395 \times 10^{-12} \text{ J}$ Convert to MeV: $1.395 \times 10^{-12} \text{ J} / (1.6 \times 10^{-19} \text{ J/eV}) / (10^6 \text{ eV/MeV}) \approx 8.7 \text{ MeV}$ Nuclear Fusion & Fission Nuclear Fusion Definition: The fusing together of two small nuclei to produce a larger nucleus. Occurs with low mass nuclei (e.g., hydrogen, helium). Requires high kinetic energy to overcome electrostatic repulsion between protons. Releases energy, as the resulting nucleus has a higher binding energy per nucleon. Nuclear Fission Definition: The splitting of a large atomic nucleus into smaller nuclei. Occurs with high mass nuclei (e.g., uranium). Induced by firing neutrons at a nucleus. Releases energy, as the daughter nuclei have higher binding energy per nucleon. Can lead to a chain reaction if ejected neutrons strike other nuclei. Used in nuclear power stations (controlled) and nuclear bombs (uncontrolled). Significance of Binding Energy per Nucleon Fusion: At low $A$, attractive nuclear forces dominate. Fusion creates a more stable nucleus with slightly less mass, releasing binding energy. Fission: At high $A$, repulsive electrostatic forces dominate. Fission converts an unstable nucleus into more stable nuclei with smaller total mass, releasing binding energy. Calculating Energy Released in Nuclear Reactions Energy released in nuclear reactions is equal to the difference in total binding energy between products and reactants. $$E_{released} = \sum E_{binding, products} - \sum E_{binding, reactants}$$ The daughter nuclei from both fission and fusion have a higher binding energy per nucleon than the parent nuclei, thus releasing energy. Worked Example: Uranium-235 fission reaction: $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{139}_{57}La + ^{95}_{42}Sr + 2^{1}_{0}n + \text{energy}$ Given binding energies per nucleon: $^{95}_{42}Sr = 8.74 \text{ MeV}$, $^{139}_{57}La = 8.39 \text{ MeV}$, $^{235}_{92}U = 7.60 \text{ MeV}$ Binding energy of $^{95}_{42}Sr = 8.74 \times 95 = 830.3 \text{ MeV}$ Binding energy of $^{139}_{57}La = 8.39 \times 139 = 1166.21 \text{ MeV}$ Binding energy of $^{235}_{92}U = 7.60 \times 235 = 1786 \text{ MeV}$ Energy released $= (830.3 + 1166.21) - 1786 = 210.51 \text{ MeV}$ Radioactive Decay The Random Nature of Radioactive Decay Radioactive decay: The spontaneous disintegration of a nucleus to form a more stable nucleus, emitting alpha, beta, or gamma particles. It is a spontaneous process : cannot be influenced by external factors (temperature, pressure, chemical conditions). It is a random process : the exact time of decay of an individual nucleus cannot be predicted, but the probability of decay is constant. The count rate of a Geiger-Muller tube exhibits fluctuations, demonstrating the random nature of decay. Activity & The Decay Constant Decay constant ($\lambda$): The probability that an individual nucleus will decay per unit time. Activity ($A$): The number of decays per unit time. Measured in Becquerels (Bq), where $1 \text{ Bq} = 1 \text{ decay s}^{-1}$. $$ A = -\frac{\Delta N}{\Delta t} = \lambda N $$ $N$ = number of undecayed nuclei remaining The negative sign indicates that $N$ decreases over time. A higher decay constant means higher activity and a faster decay rate. Worked Example: Americium-241 (mass $5.1 \text{ µg}$) has an activity of $5.9 \times 10^5 \text{ Bq}$. Molecular mass = 241. Number of nuclei ($N$): $$ N = \frac{\text{mass} \times N_A}{\text{molecular mass}} = \frac{(5.1 \times 10^{-6} \text{ g}) \times (6.02 \times 10^{23} \text{ mol}^{-1})}{241 \text{ g/mol}} \approx 1.27 \times 10^{16} $$ Decay constant ($\lambda$): $$ \lambda = \frac{A}{N} = \frac{5.9 \times 10^5 \text{ Bq}}{1.27 \times 10^{16}} \approx 4.65 \times 10^{-11} \text{ s}^{-1} $$ The Exponential Nature of Radioactive Decay The number of undecayed nuclei ($N$) decreases exponentially over time. $$ N = N_0 e^{-\lambda t} $$ $N_0$ = initial number of undecayed nuclei $t$ = time interval Activity ($A$) and count rate ($C$) also follow an exponential decay: $$ A = A_0 e^{-\lambda t} $$ $$ C = C_0 e^{-\lambda t} $$ $e$ is the exponential constant (approx. 2.718). Its inverse function is the natural logarithm, $\ln$. Worked Example: Strontium-90 has $\lambda = 0.025 \text{ year}^{-1}$. Determine activity $A$ after 5 years as a fraction of initial activity $A_0$. $$ \frac{A}{A_0} = e^{-\lambda t} = e^{-(0.025 \text{ year}^{-1}) \times (5.0 \text{ years})} = e^{-0.125} \approx 0.88 $$ The activity decreases to 88% of its initial value. Half-Life Half-life ($t_{1/2}$): The time taken for the initial number of nuclei (or activity) to reduce by half. Derivation from $N = N_0 e^{-\lambda t}$: When $t = t_{1/2}$, $N = N_0/2$. $$ \frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}} $$ $$ \frac{1}{2} = e^{-\lambda t_{1/2}} $$ Taking natural logarithm of both sides: $$ \ln \left(\frac{1}{2}\right) = -\lambda t_{1/2} $$ $$ -\ln(2) = -\lambda t_{1/2} $$ $$ t_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.693}{\lambda} $$ Half-life and decay constant are inversely proportional: shorter half-life means larger decay constant and faster decay. Worked Example: Strontium-90 has a half-life of 28.0 years. Calculate $\lambda$ in $\text{s}^{-1}$. Convert half-life to seconds: $t_{1/2} = 28.0 \text{ years} \times 365 \text{ days/year} \times 24 \text{ hours/day} \times 60 \text{ min/hour} \times 60 \text{ s/min} \approx 8.83 \times 10^8 \text{ s}$ $$ \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{8.83 \times 10^8 \text{ s}} \approx 7.85 \times 10^{-10} \text{ s}^{-1} $$