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The intensity field can in principle be solved from the integrodifferential radiative transfer equation (RTE), but an exact solution is usually impossible and even in the case of geometrically simple systems can contain unusual special functions such as the Chandrasekhar's H-function and Chandrasekhar's X- and Y-functions. [3]
There are a wide variety of fields associated with nuclear engineering, but computers and associated software are used most often in design and analysis. Neutron kinetics, thermal-hydraulics, and structural mechanics are all important in this effort. Each software needs to be tested and verified before use. [1]
The boundaries of the valley of stability, that is, the upper limits of the valley walls, are the neutron drip line on the neutron-rich side, and the proton drip line on the proton-rich side. The nucleon drip lines are at the extremes of the neutron-proton ratio. At neutron–proton ratios beyond the drip lines, no nuclei can exist.
Defining equation SI units Dimension Number of atoms N = Number of atoms remaining at time t. N 0 = Initial number of atoms at time t = 0 N D = Number of atoms decayed at time t = + dimensionless dimensionless Decay rate, activity of a radioisotope: A = Bq = Hz = s −1 [T] −1: Decay constant: λ
The neutron transport equation is a balance statement that conserves neutrons. Each term represents a gain or a loss of a neutron, and the balance, in essence, claims that neutrons gained equals neutrons lost. It is formulated as follows: [1]
Comparison between the Nuclear Force and the Coulomb Force. a – residual strong force (nuclear force), rapidly decreases to insignificance at distances beyond about 2.5 fm, b – at distances less than ~ 0.7 fm between nucleons centres the nuclear force becomes repulsive, c – coulomb repulsion force between two protons (over 3 fm, force becomes the main), d – equilibrium position for ...
The first step of the proton-proton chain is a two-stage process: first, two protons fuse to form a diproton: 1 H + 1 H + 1.25 MeV → 2 He; then the diproton immediately beta-plus decays into deuterium: 2 He → 2 H + e + + ν e + 1.67 MeV; with the overall formula 1 H + 1 H → 2 H + e + + ν e + 0.42 MeV.
Therefore, a nucleus with an even number of protons and an even number of neutrons has 0 spin and positive parity. A nucleus with an even number of protons and an odd number of neutrons (or vice versa) has the parity of the last neutron (or proton), and the spin equal to the total angular momentum of this neutron (or proton).