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Figure I S \how\ a MOSFET', witching ~nterval, with the upper graph being the voltage and current on the Drain IU Source and the I w x r graph detailing Vti9 v\ tiiiie with a coii\tilnt current charging the gate The x-axi, therefore i\ alw repre*cntative of gate charge (Qcl CIFs= CGD CGF, and 11 contruk t I, 0. and 14 timing. CGDreceive\ the current trum the gate driver during 13 ( 8 , V I\ fallmgJ. The gate D I charge (QaJpariimeteri on the lower graph are either \prcitied or c be drmed from MOSFET datasheet\ m

TABLE 29-1

You can examine multi-value properties in Listing 15-4. We ve used a side-by-side code example so you can make a direct comparison of safe array-handling techniques in VBScript and JScript.

where n is the solvent index. In electrolyte systems, the electroneutrality condition has to be met in each point of the liquid phase:

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If we denote R as r0 A1/3, we can substitute for R in the above equation, arriving at the point that ECoul 0.72 Z 2/A 1/3 MeV. (The usual tted or adjusted value of the coef cient ac is 0.7 rather than 0.72.) 4. The fourth term along with the fth term represent effects on the binding energy that are quantum mechanical in origin. The fourth term, the asymmetry energy, is the difference in energy of a nucleus with N neutrons and Z protons (N = Z) and one where Z N A/2. To evaluate this term, we remember that neutrons and protons occupy orbitals in the nucleus at well-de ned energies and that the neutrons and protons obey the Pauli principle for fermions. A simple model (Fig. 2.4) should suf ce to calculate this energy. Assume the neutron and proton levels of a nucleus are equidistant with spacing D and that we can have only one nucleon per level. To make the nucleus AZ (where Z = N) from the nucleus with N Z A/2, we must take q protons and transform them into neutrons. Thus, we have N q A/2, Z A/22q and therefore q (N 2 Z )/2. Each of the q protons must be raised in energy an amount qD. The work needed to transform the N Z nucleus into the nucleus AZ is q 2D (N 2 Z )2D/4. Note that we could have made exactly the same argument by replacing neutrons with protons. We nish the argument by noting that the

Let us now consider the case where G=D ) 1, that is, many overlapping levels of the compound nucleus are populated (We are also tacitly assuming a large range of compound nuclear excitation energies) The cross section for the reaction a A ! C ! b B can be written as sab sC PC (b) where sC (a) is the cross section for the formation of the compound nucleus C and PC is the probability that C will decay to form b B Clearly SPC (b) 1 Now let us consider, in detail, the probability that emitted particle b has an energy 1b First of all, we can write down that the maximum energy that b can have is EC 2 Sb where EC is the excitation energy of the compound nucleus and Sb is the separation energy of b in the residual nucleus B.

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But b can be emitted with a variety of energies less than this with the result that the nucleus B will be left in an excited state By using the arguments of detailed balance from statistical mechanics (see Lefort, 1968 or Friedlander, Kennedy, Miller, and Macias (FKMM)) we can write for the probability of emitting a particle b with an energy 1b (, 1max, leaving the nucleus B at an excitation energy EB) r EB (2Jb 1)m Wb 1b d1b 1b sinv d1b r EC p2 h 3 In this equation, m is the reduced mass of the system, and sinv is the cross section for the inverse process in which the particle b is captured by the nucleus B where b has an energy, 1b.

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