Thermodynamic Stability of the Nucleus

After consumption of Na l, the patient s thyroid is scanned for radioactivity levels to determine the efficiency of iodine absorption, (left) A normal thyroid, (right) An enlarged thyroid. 

Nuclide Half-Life Area of the Body Studied 

99mTc 6.0 hours Heart, bones, liver, and lungs 

Radiotracers provide sensitive and noninvasive methods for learning about biologic systems, for detection of disease, for monitoring the action and effectiveness of drugs, and for early detection of pregnancy, and their usefulness should continue to grow. Some useful radiotracers are listed in Table 19.5. 

We can determine the thermodynamic stability of a nucleus by calculating the change in potential energy that would occur if that nucleus were formed from its constituent protons and neutrons. For example, let s consider the hypothetical process of forming a eO nucleus from eight neutrons and eight protons  

The m in technetium-99m designates an excited nuclear state of Tc that decays to the ground state by y production  

TABLE 21.5 Some Radioactive Nuclides, with Half-Lives and Medical Applications as Radiotracers 

A relatively new technique that uses radioactivity to study body processes and diagnose malfunctions is commonly called positron emission tomography (PET). In this technique radionuclides that decay by positron emission are incorporated into compounds. For example, brain function can be studied by incorporating 1gC into glucose, which is the main source of energy for the brain. By studying how this labeled glucose is metabolized in the brain, doctors can discover abnormalities caused by diseases such as cancer, Parkinson s disease, and epilepsy. 

The energy change associated with this process can be calculated by comparing the sum of the masses of eight protons and eight nentrons with that of the oxygen nucleus  

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The energy changes associated with normal chemical reactions are small enough that the corresponding mass changes are not detectable. 

The difference in mass for formation of 1 mole of nuclei is therefore 

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Here, — T, defined as AT, is the undercooling. Thermodynamic stability of the nucleus is attained when AG = 0, and... 

This may continue until eventually the cluster is large enough to be thermodynamically stable (i.e., will not redissolve). However, if the cluster is smaller than the critical nucleus size, then there is the possibility that the nucleus will redissolve. The lifetime of the nucleus will then depend on its size and also on the temperature lower temperatures will slow the redissolution step. Thus lower temperature increases the chance that a subcritical nucleus will eventually grow to a stable size rather than redissolve. This kinetic stabilization of small nuclei results in a greater total density of nuclei and therefore smaller crystal size, since the total quantities of reactants are fixed. 

The thermodynamic stability of a particular nucleus is normally represented as energy released per nucleon. To illustrate how this quantity is obtained, we will continue to consider gO. First, we calculate AE for one mole of gO nuclei from the equation... 

Thermodynamic stability (nuclear) the potential energy of a particular nucleus as compared with the sum of the potential energies of its component protons and neutrons. (21.1) Thermodynamics the study of energy and its interconversions. (9.1)... 

We treat, in this chapter, mainly solid composed of water molecules such as ices and clathrate hydrates, and show recent significant contribution of simulation studies to our understanding of thermodynamic stability of those crystals in conjunction with structural morphology. Simulation technique adopted here is not limited to molecular dynamics (MD) and Monte Carlo (MC) simulations[l] but does include other method such as lattice dynamics. Electronic state as well as nucleus motion can be solved by the density functional theory[2]. Here we focus, however, our attention on the ambient condition where electronic state and character of the chemical bonds of individual molecules remain intact. Thus, we restrict ourselves to the usual simulation with intermolecular interactions given a priori. 

Why do the adduct carbocations 46 not react with a nucleophile to give an adduct of type 48, as happens in electrophilic addition to double bonds The main reason is probably thermodynamic. The hypothetical reaction (5.32) is very exothermic, based on experimental heats of formation. If we attribute this exothermicity to the stabilization of the benzene nucleus, we obtain a value of 150 kJ mol-1 for this stabilization. This stabilization would be lost in reaction (5.31) if the adduct 48 were formed, but is regained if the substituted product 47 is produced. This stabilization does not apply to reaction with alkenes, so addition can take place if the reaction is favourable energetically. 

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