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烟台大学硕士研究生20 13 ～20 14 学年第 1 学期
Consider a Pb-Sn solder that is 38.1wt% Pb and 61.9wt% Sn (this is the eutectic composition with the lowest melting point). What are the atomic fractions of Pb and Sn in this solder? Given MPb=207.2 and MSn=118.71.
Problem 2 (resistivity of intrinsic and doped Si. 20 points)
Find the resistance of a 1 cm3 pure silicon crystal cube. What is the resistance when the crystal is doped with arsenic (As) if the doping is 1 in 109, that is, 1 part per billion (ppb) (note that this doping corresponds to one foreigner living in China)? Given data: atomic concentration in silicon is 5?1022cm-3, ni=1.0?1010cm-3, ?e=1350cm2V-1s-1, and ?h=450cm2V-1s-1. If doping with boron instead of arsenic (a p-type semiconductor with the same level of doping.
a. A 50g golf ball traveling at a velocity of 20ms-1.
b. A proton traveling at 2200mS-1. Given mp=1.67?10-27kg.
c. An electron accelerated by 100V. Given me=9.10938188 × 10-31. Consider two copper wires separate only by their surface oxide layer (CuO). Classically, since the oxide layer is an insulator, no current should be possible through the two copper wires. Suppose that for the conduction (“free”) electrons in copper, the surface oxide layer looks like a square potential energy barrier of height 10eV. Consider an oxide layer thickness of 5 nm and evaluate the transmission coefficient for conduction electrons in copper, which have a kinetic energy of about 7 eV. What will be the transmission coefficient if the oxide barrier is 1 nm?
Problem 5 (Noise in RLC circuit. 10 points)
Most radio receivers have a tuned parallel-resonant circuit, which consist of an inductor L, capacitor C and resistance R in parallel. Suppose L is 100?H; C is 100pF; and R, the equivalent resistance due to the input resistance of amplifier and to the loss in the coil (coil resistance plus ferrite losses), is about 200k?. What is the minimum rms (root mean square) radio signal that can be detected? we develop a basic understanding of the properties of intrinsic and extrinsic semiconductors. Although most of our discussions and examples will be based on Si, the ideas are applicable to Ge and to the compound semiconductors such as GaAs, InP, and others. By intrinsic Si we mean an ideal perfect crystal of Si that has no impurities or crystal defects such as dislocations and grain boundaries. The crystal thus consists of Si atoms perfectly bonded to each other in the diamond structure. At temperatures above absolute zero, we know that the Si atoms in the crystal lattice will be vibrating with a distribution of energies. Even though the average energy of the vibrations is at most 3kT and incapable of breaking the Si-Si bond, a few of the lattice vibrations in certain crystal regions may nonetheless be sufficiently energetic to “rupture” a Si-Si bond. When a Si-Si bond is broken, a “ free ”electron is created that can wander around the crystal and also contribute to electrical conduction in the presence of an applied field. The broken bond has a missing electron that causes this region to be positively charged. The vacancy left behind by the missing electron in the bonding orbital is called a hole.
An electron in a neighboring bond can readily tunnel into this broken bond and fill it, thereby effectively causing the hole to be displaced to the original position of the tunneling electron. By electron tunneling from a neighboring bond, holes are therefore also free to wander around the crystal and also contribute to electrical conduction in the presence of an applied field. In an intrinsic semiconductor, the number of thermally generated electrons is equal to the number of holes (broken bonds). In an extrinsic semiconductor, impurities are added to the semiconductor that can contribute either excess electrons or excess holes. For example, when an impurity such as arsenic (As) is added to Si, each As atom acts as a donor and contributes a free electron to the crystal. Since these electrons do not come from broken bonds, the numbers of electrons and holes are not equal in an extrinsic semiconductor, and the As-doped Si in this example will have excess electrons. It will be an n-type Si since electrical conduction will be mainly due to the motion of electrons. It is also possible to obtain a p-type Si crystal in which hole concentration is in excess of the electron concentration due to, for example, boron (B) doping.