=
Z
Ψ∗ (x, t)(−i¯ h
= −i¯ h
Z
ψ(x)
∂ ∗ [Ψ (x, t)]dx ∂x
∂ [ψ(x)]dx ∂x
and the integral given in the hint, we find that the average momentum < p > is zero. (e) (4 pts.) Qualitatively compare the probabilities of finding the particle in the region 0.95 L < x < 1.05 L for the particle in the ground state (n = 1) and the next higher state (n = 2). (Hint: Sketch the wavefunctions and consider the probability density functions for these two states.)
Comparing the area for the probabilty density functions (|ψ(x)|2 ) for these two states in the figure on the next page, it is clear that the probability for the n = 2 state for finding the particle in the region 0.95 L < x < 1.05 L is higher: (f) (BONUS: 5 pts.) We know that the particle in any of the states of the infinite potential well has to have a nonzero kinetic energy, and therefore nonzero momentum. Why is the result of part d) not inconsistent with this statement? 4
The particle in any state will have a positive kinetic energy and therefore nonzero momentum; however, it is equally likely to be moving in the −ˆ x direction as in the xˆ direction. Therefore the average momentum, if you look at many particles in this state of many wells, comes out to zero.
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PART III—CARRIERS IN SEMICONDUCTORS, 35 pts. 3. In this question, consider two semiconductor materials, Aa and Bb. The band-symmetry approximation holds for both materials; that is to say, NC,Aa = NV,Aa and NC,Bb = NV,Bb where NC and NV are the effective density of states for the conduction and valence bands, respectively. (a) (5 pts.) Assume NC,Aa = 1.5NC,Bb and Eg,Aa = 0.5Eg,Bb , where Eg is the bandgap energy of the material. At the same temperature T1 , which material has the higher intrinsic carrier concentration ni ? Show your reasoning.
Since NC = NV , we can write the intrinsic carrier concentration for either semiconductor as q −Eg NC2 exp ( ) ni = 2kT −Eg ) = NC exp ( 2kT Then, for these materials: −0.5Eg,Bb ni,Aa = 1.5NC,Bb exp ( ) 2kT −Eg,Bb = 1.5NC,Bb exp ( ) 4kT −Eg,Bb ni,Bb = NC,Bb exp ( ) 2kT ⇒ ni,Aa > ni,Bb This makes sense, since Aa has both a higher density of states than Bb, and a smaller bandgap that makes it easier for EHPs to be generated. (b) (5 pts.) Now assume NC,Aa = NC,Bb and Eg,Aa = 0.5Eg,Bb . If the temperature is raised to 2T1 , which material experiences the larger change in ni ? Show your reasoning.
Define NC = NC,Aa = NC,Bb . Then you can calculate the intrinsic carrier concentration for each material at T1 and 2T1 and compare. It seems that the ratio of ni at 2T1 to that at T1 is higher for Bb; however if you look at the absolute change |ni (2T1 ) − ni (T1 )|, ni,Aa changes more by orders of magnitude, since its lower bandgap makes its carrier concentration and bandgap much more sensitive to temperature changes. 4. Consider a silicon sample doped uniformly with donor atoms, so that ND = 2 × 1016 1/cm3 . Use Eg =1.12 eV for the bandgap. (a) (10 pts.) Find n0 , p0 and the location of EF in the bandgap at room temperature (300 ◦ K) at equilibrium. Sketch the band energy diagram; mark EC , EV , Ei and EF .
The sample only has donor atoms; therefore it is n-type. So: n0 ≈ ND = 2 × 1016 1/cm3 n2i p0 = = 1.05 × 104 1/cm3 n0 kT n0 (Ei − EF ) = ln ( ) = 0.368 eV q ni 6
(b) (8 pts.) Assume the mean-free time τf is the same for electrons and holes. Assume that the effective mass of holes is three times that of electrons, and assume µn =1500 cm2 /V.sec is the electron mobility. What are the conductivity due to electrons and the conductivity due to holes at equilibrium at 300 ◦ K? Write the net drift current density in terms of an applied electric field E.
m∗n = m∗p /3 ⇒ µp = µn /3 = 500 cm2 /V.sec σn = qµn n0 = 1.6 × 10−19 × 1500 × 2 × 1016 = 4.8 1/Ω.cm σp = 1µp p0 = 1.6 × 10−19 × 500 × 10500 = 8.4 × 10−13 1/Ω.cm Jnet,drif t = Jn,drif t + Jp,drif t ≈ Jn,drif t = σn E = 4.8E A/cm2 since in this case we can almost ignore the hole drift current with respect to the electron drift current. (c) (7 pts.) Suggest a possible acceptor density NA to add to this sample so that the conductivities due to both carrier types become comparable (for instance, the lesser is no less than 1/100th of the greater). Show your work.
Let’s start by choosing a target conductivity; for example, σp = σn /30. Because of the mobility ratio, this implies that we need p = 3n: The new material will be p-type. Thus we have the following equations to solve for p and n, and thus for NA : p = 3n n2 n= i p ⇒p ⇒ NA
= NA − ND n2 ni ⇒ n = i ⇒ n = √ = 8.37 × 109 1/cm3 3n 3 10 3 = 2.15 × 10 1/cm = ND + 2.15 × 1010 = 2.00000215 × 1016 1/cm3
So to get the minority and majority carrier concentrations to be comparable, which is required to get the minority and majority conductivities to be comparable, the added acceptor density needs to be quite close to the added donor density. 7
Part IV—15 pts. 5. Fill in the blanks, choose the correct statement or indicate “True” or “False” for the statements below. Each blank, true/false or multiple choice answer is worth one point. Blanks might contain more than one word. You may choose 15 to answer; if you get more than 15 answers right in this section your grade will be 15. (a) Increasing the intensity of light falling on a metal increases the number of photoelectrons emitted, but only if the frequency the light is high enough. (b) One of Bohr’s postulates implies that the diameter of a stable orbit around the atomic nucleus has to be an integer multiple of the associated wavelength OR de Broglie wavelength of the electron there. (c) Electrons in a semiconductor crystal contribute to conduction only if their energy level falls into the conduction band.. (d) Multiple Choice: In Part I, Question 1 of this test, the crystal surface with the highest number of type A atoms in the unit cell volume is— (a) (010), (b) (110), (c) (111) . (e) True/False: Blue light at λ = 250 nm can create electron-hole pairs in intrinsic silicon carbide, with bandgap 3.26 eV: TRUE (f) True/False: If a p-type material is doped with additional acceptors, the Fermi level gets closer to the conduction band: FALSE (g) A crystal can be formed by placing a basis (an arrangement of atoms) at each lattice point. (h) If the curvature of the E −k relationship of the conduction band of a material in a given direction in the crystal is gentler than the curvature in another direction, the electron effective mass is higher in the first direction. (i) Acceptors are impurities that have a free bond OR valence state OR state able to accept an electron when placed in a semiconductor lattice. (j) True/False: In a p-type semiconductor, (1 − f (EV )) < f (EC ): FALSE (k) In an extrinsic semiconductor doped with acceptors, electrons are the minority carriers. (l) Rate of recombination increases with increasing carrier concentration if the system is not at equilibrium, but if no external sources are present, rate of generation stays constant as long as temperature is unchanged. (m) Multiple choice Which one of the following is a valid (n, l, m) triplet for the quantum numbers of an electron bound to an atom? (a) (2,3,0) (b) (1,0,-1) (c) (4,2,-2) (d) (3,3,-1) (n) Multiple choice The energy of an electron in orbit around the hydrogen atom comes from the solution to the ......-dependent equation obtained by using separation of variables on the Schroedinger equation. (a) θ (b) φ (c) r (o) The diamond lattice has 4 lattice points that are enclosed entirely within the unit cell.
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