几道半导体物理材料题（英文）求解,Useful constants:kB = 1.38× 10-23 J/K,c = 2.998 x 108 m/s,h= 6.626 x 10-34 J-s m= 9.109 x 10-31 kg,e = 1.602 x10-19 C1.Materials in the nanometer scale exhibit physical properties quite different from

几道半导体物理材料题（英文）求解,Useful constants:kB = 1.38× 10-23 J/K,c = 2.998 x 108 m/s,h= 6.626 x 10-34 J-s m= 9.109 x 10-31 kg,e = 1.602 x10-19 C1.Materials in the nanometer scale exhibit physical properties quite different from
几道半导体物理材料题（英文）求解,
Useful constants:kB = 1.38× 10-23 J/K,c = 2.998 x 108 m/s,h= 6.626 x 10-34 J-s
m= 9.109 x 10-31 kg,e = 1.602 x10-19 C
1.Materials in the nanometer scale exhibit physical properties quite different from bulk materials.
a.Crystals in the nanometer scale have a lower melting temperature and the difference can be as large as 1000 °C.Why?
b.Surface energy (γ) can be defined as the energy required to create a unit area of “new” surface.Considering that two new surfaces can be crated by separating solid materials into two pieces,define the surface energy of {100} surface in FCC crystal?Here the atomic bond strength is ε.
c.The surface energies calculated from (b) exhibit some deviation from experimental data,and this is can be understood by the surface energy reduction mechanism.Name 4 ‘surface energy reduction mechanisms’.
3.[28 points] As shown in Fig.1, an electron is trapped in a one-dimensional energy well of width L with rigid walls. Here, time-independent Schrödinger equation that can describe the motion of electrons is given by,
and the general wave solution in region II is given by, 
a. Determine k, A2, and B2. 
b. Determined the allowed energy levels (En) in Region II. 
c. For L = 1 nm, calculate wavelengths of the two lowest-energy photons capable of exciting electrons from the ground state. 
d. Calculate the maximum well width, L that can ensure us the observation of electron transition between the two lowest energy levels at T = 300K.
e. If photoluminescent (PL) spectra from InP microcrystals and InP nanowire (diameter, d = 50 nm) exhibit the dominant near band-edge (NBE) peaks at 1.45 eV, and 1.47 eV, respectively. Then, estimate the energy of NBE peak from 10 nm thick InP nanowire? (Ignore the Coulomb interaction.)

几道半导体物理材料题（英文）求解,Useful constants:kB = 1.38× 10-23 J/K,c = 2.998 x 108 m/s,h= 6.626 x 10-34 J-s m= 9.109 x 10-31 kg,e = 1.602 x10-19 C1.Materials in the nanometer scale exhibit physical properties quite different from
For Q3,Where is Fig 题目给清楚,补充问题我帮你解.其他的自己维基百科搜吧