Fluid Mechanics

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Fluid Mechanics

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Homework 5 Solution
Question 14.6 Define the following quantities pertaining to the flow of electrons in a metal conductor: (a) drift velocity; (b) relaxation time; (c) electron mobility. Answer 14.6 (a) The drift velocity vd, is the average velocity achieved by an electron in the presence of a uniform electric field, E. An average value is used since the electron??™s motion varies periodically, in a sawtooth manner, as the particle accelerates, collides with positive-ion cores, and reaccelerates. (b) The relaxation time ?, is the average time between collisions of a conduction electron with the positive-ion cores of the metal lattice. (c) The electron mobility, ?µ, is the proportionality constant relating to the drift velocity to the applied electric field as vd = ?µ*E. The SI units for ?µ are m2/(V*s). Question 14.8 What structural defects contribute to the residual component of the electrical resistivity of a pure metal Answer 14.8 Dislocations, vacancies, grain boundaries and impurity atoms are common structural defects that contribute to the small residual component of a pure metal??™s electrical resistivity. Question 14.13 What explanation is given for the good electrical conductivity of magnesium and aluminum even though these metals have filled outer 3s energy bands Answer 14.13 Although the outer 3s energy bands in magnesium and aluminum are filled, these metals have good electrical conductivity because their 3s bands overlap their 3p bands. In the case of magnesium, the empty 3p band combines with the 3s band to form a partially filled 3sp band. Similarly, in aluminum, the 3p band, which contains one electron, overlaps the full 3s band. Question 14.14 How does the energy-band model explain the poor electrical conductivity of an insulator such as pure diamond Answer 14.14 In an insulator, such as pure diamond, the electrons are tightly bound in covalent or ionic bonds. The energy-band model theorizes that these bound electrons fill a lower valence band which is separated by a large energy gap, Eg, from an empty outer conduction band. A large potential is thus necessary for an electron to overcome the gap. In the case of pure diamond, approximately 6 to 7 eV are required to free an electron for conduction. Question 14.15 Define an intrinsic semiconductor. What are the two most important elemental semiconductors Answer 14.15 An intrinsic semiconductor is a pure semiconductor whose electrical conductivity is a function of the temperature and inherent conductive properties of the material, such as its energy gap and bonding structure. Two of the most important elemental semiconductors are Si and Ge.

Question 14.21 Define n-type and p-type extrinsic silicon semiconductors. Answer 14.21 An n-type (negative-type) extrinsic silicon semiconductor is a semiconducting material that was produced by doping silicon with an n-type element of Group V A, such as P, As, or Sb. Consequently, electrons are the majority charge carriers of the material. A p-type (positive-type) extrinsic silicon semiconductor is a semiconducting material that was produced by doping silicon with an p-type element of Group IIIA, such as B, Al, or Ga. Since the dopants are acceptor atoms, holes are the majority charge carriers of the material. Question 14.22 Draw energy-band diagrams showing donor or acceptor levels for the following: (a) n-type silicon with phosphorus impurity atoms; (b) p-type silicon with boron impurity atoms. Answer 14.22

Question 14.27 Describe the movement of majority carriers in a pn junction diode at equilibrium. What is the deletion region of a pn junction Answer 14.27 In a pn junction diode at equilibrium, there is no movement of majority carriers due to the potential difference established within the depletion region. This depletion region, formed at the junction of the n-type and p-type semiconductors by the diffusion and recombination of majority carriers, consists of large heavy negatively charged ions on the p-type border and positively charged ions on the n-type border. The opposing charges create an electrical potential and repel the motion of majority carriers. Question 14.28 Describe the movement of the majority and minority carriers in a pn junction diode under reverse bias.

Answer 14.28 Under reverse bias, majority carriers move away from the pn junction, and thus increase the depletion width, while minority carriers flow toward the junction and create a very small leakage current on the order of microamperes. Question 14.61 Calculate the electrical resistivity (in ohm-meters) of a silver wire 15 m long and 0.030 m in diameter at 160?°C. [?e(Fe at 0?°C) = 9.0 ? 10-6 ? m] Answer 14.61 The electrical resistivity is independent of the wire length and diameter but dependent on temperature. From Table 13.3, the resistivity for silver at 0?°C is 1.47?10-6 ??cm. Thus, at 160?°C,

Question 14.63 At what temperature will the electrical resistivity of an iron wire be 25.0 ?10-8 ?*m Answer 14.63 ? t = ? 0 (1+ ? T T)
25 ?10?8 = (9 ?10?6 ) * (1+ 0.0045T) T = 395?°C

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Question 14.65 What is the ratio of the electron-to-hole mobility in silicon and germanium Answer 14.65

Question 14.67 Calculate the electrical resistivity of germanium at 300K. Answer 14.67

Question 14.68 The electrical resistivity of pure germanium is 0.46 ??m at 300K. Calculate its electrical conductivity at 425?°C. Answer 14.68

Question 14.72 A semiconductor is made by adding boron to silicon to give an electrical resistivity of 1.90 ??m. Calculate the concentration of carriers per cubic meter in the material. [Assume ?µp = 0.048 m2 /(V*s).] Answer 14.72 For this p-type semiconductor, the concentration of holes per cubic meter is:

Question 14.77 Describe the origin of the three stages that appear in the plot of ln ? versu 1/T for an extrinsic silicon semiconductor (going from low to high temperatures). Why does the conductivity decrease just before the rapid increase due to intrinsic conductivity Answer 14.77