Summary

This document provides an explanation of radiochemistry concepts, including radioactive decay, decay constants, half-life, and atomic disintegration. It also features worked examples and problems related to these topics. The document is suitable for undergraduate-level study.

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# Radiochemistry Radiochemistry is the branch of chemistry that deals with radioactive materials and their properties. It involves the study of the chemical properties and behavior of elements, isotopes and compounds that exhibit radioactivity. ## Isotopes Atoms having the same number of protons...

# Radiochemistry Radiochemistry is the branch of chemistry that deals with radioactive materials and their properties. It involves the study of the chemical properties and behavior of elements, isotopes and compounds that exhibit radioactivity. ## Isotopes Atoms having the same number of protons but a different number of neutrons are isotopes. ## Radioactivity Radioactivity is the spontaneous emission of radiation, often in the form of alpha particles, Beta particles, or gamma particles by unstable atomic nuclei undergoing nuclear decay. ## Decay constant The decay constant, denoted by the symbol λ, is a parameter used in the context of radioactive decay. It represents the probability per unit time that a radioactive nucleus will decay. ### Derivation Let's consider the simple radioactive decay process, where the rate of change of the number of radioactive nuclei (N) with respect to time (t) is proportional to the number of nuclei present. $ \frac{dN}{dt} = -λN $ This indicates, rate decay α current no. of radioactive nuclei and the negative sign indicates, the number of nuclei is decreasing over time. On integrating: $∫\frac{1}{N} dN =-λ ∫dt$ ⇒ ln|N| = -λt + C Now, at t = 0, N(0) = No, so, ln|No| = C Put the value of C in the aforementioned equation: ln|N| = -λt + ln|No| ⇒ ln($\frac{N}{No}$) = -λt ⇒ $\frac{N}{No}$ = e^-λt ⇒ N(t) = No ⋅ e^-λt where, λ = -2.303 log $\frac{N}{No}$ / t ## Half-life The amount of time required for half of the radioactive atoms to disintegrate. For first order kinetics, half-life is: t<sub>1/2</sub> = 0.693 / λ ## Atomic Disintegration The rate of atomic disintegration or radioactive decay is measured by the activity of a radioactive substance. Activity is the number of decays per unit time and is expressed in becquerels (Bq) or curies (Ci). ### Formula for radioactive decay or atomic disintegration A(t) = Ao ⋅ e<sup>-λt</sup> where, * A(t) = quantity of radioactive substance at time t * Ao = initial quantity of radioactive substance λ is decay constant, which is related to the half-life (T<sub>1/2</sub>) by λ = ln(2) / T<sub>1/2</sub> ∴ A(t) = Ao e<sup>-λt</sup> = Ao (1/2)<sup>t/T<sub>1/2</sub></sup> and rate of atomic disintegration = λN ## Problem 1 The amount of C-14 in a piece of wood is found to be one-sixth of its amount in fresh piece of wood. Calculate the age of one piece of wood. Half-life of C-14 is 5,730 years. Given that, ratio of amount of C-14 in the old piece of wood to the amount of C-14 in fresh piece of wood by, At/Ao = 1/6 Radioactive decay At = Ao(1/2)<sup>t/T<sub>1/2</sub></sup> ⇒ 1/6 = (1/2)<sup>t/5730</sup> Taking log<sub>2</sub> both sides: log<sub>2</sub>(1/6) = t / 5730 ⇒ t = 5730 . log<sub>2</sub>(1/6) ## Problem 2 A bone taken from a garbage pile buried under a hillside had a C-14/C-12 ratio 0.477 times the ratio in a living animal. Calculate the time in years of this burial. Half-life of C-14 is 5,730 years. The ratio of C-14 to C-12 in a living organism is assumed to be a constant. The ratio of C-14 to C-12 (R) in a sample can be related to the elapsed time since the death of the organism using the formula: R = Ro (1/2)<sup>t/T<sub>1/2</sub></sup> Given that R = 0.477×Ro (put in the above formula) 0.477×Ro = Ro(1/2)<sup>t/T<sub>1/2</sub></sup> 0.477 = (1/2)<sup>t/5730</sup> Taking log<sub>2</sub> both sides: log<sub>2</sub>(0.477) = t/5730 ⇒ t = 5730 . log<sub>2</sub>(0.477) ## Radioactive Equilibrium (Not so important) Radioactive equilibrium occurs when the rate of production of a radioactive daughter isotope is equal to its rate of decay. In other words, the concentration of the parent and daughter isotopes remains constant over time. Let, * Rp be the rate of production of parent isotope * Rd be the decay rate of daughter isotope Then, the equilibrium condition can be expressed as: Rp = Rd **Example:** The uranium decay chain, uranium-238 (<sup>238</sup>U) decays into thorium-234 (<sup>234</sup>Th), which then undergoes further decay. If the rate of production of thorium-234 matches its rate of decay, an equilibrium is reached where the concentrations of uranium-238 and thorium-234 remain relatively constant. ## Radioisotopes and their applications Radioisotopes are unstable isotopes of elements that undergo radioactive decay, emitting radiation in the form of alpha particles, beta particles and gamma rays. ### Applications: * Radioisotopes are largely used in nuclear medicine for diagnostic imaging and cancer treatment (e.g., iodine-131 in thyroid cancer therapy). * They can be used as tracers to trace the movement of substances within the body. * Radioisotopes are utilized in agriculture for studying plant growth and investigating the movement of substance in soil, e.g. Phosphorous-32 is used as a tracer in plant studies. * Carbon dating, based on the decay of C-14 is used to determine the age of archaeological artifacts and fossils. * Some radioisotopes such as U-235 and plutonium-239 are used as fuel in nuclear reactors. ## Theories behind the stability of heavy nucleus of elements 1. **Nuclear Shell model:** * Describes the nucleus as having energy levels or shells for protons and neutrons similar to electron shells in atoms. * Nuclei with 'magic number' of protons or neutrons (such as 2, 8, 20, 28, 50, 82, and 126) are considered more stable. 2. **Liquid Drop Model:** * According to this model, the nucleus experiences attractive nuclear forces that resist Coulomb repulsion between protons. * The balance between these forces contributes to stability. 3. **Fission and Fusion:** * **Nuclear fusion** is a process in which two light atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. The fusion process requires extremely high temperatures and pressures to overcome the electrostatic repulsion between two charged atomic nuclei. An example of fusion reaction is the combination of deuterium (D<sub>2</sub>) and tritium (T<sub>3</sub>) to form helium (He<sub>4</sub>) and a neutron (n<sub>0</sub>): D<sub>2</sub> + T<sub>3</sub> → He<sub>4</sub> + n<sub>0</sub> + Energy * **Nuclear fission** is the process in which a heavy atomic nucleus splits into two or more lighter nuclei. In a nuclear fission reaction, a heavy nucleus, often U-235 (<sup>235</sup>U) or plutonium-239 (<sup>239</sup>Pu), absorbs a neutron, becoming unstable and then splitting into two or more lighter nuclei releasing neutrons and energy. Example: <sup>235</sup>U<sub>92</sub> + n<sub>0</sub> → <sup>141</sup>Ba<sub>56</sub> + <sup>92</sup>Kr<sub>36</sub> + 3n<sub>0</sub> + Energy Released neutrons go on to further fission reactions.

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