HONG KONG BAPTIST UNIVERSITY
SEMESTER II EXAMINATION, 1998-99

Subject Code: SCI 3421/2                     Section No.: 00001                           Time Allowed : 3 Hour(s)

Subject Title:   Material Science: Thin Films, Semiconductors I/II
Solutions
1. The thinner of foil, the higher the concentration gradient, so the greater will be the diffusional flux J.

Fick’s first law:  where D is the diffusivity of hydrogen in palladium, c the concentration of hydrogen, and xo the thickness of palladium foil.
 
2. The higher the acceleration voltage V, the bigger the kinetic energy of each electron E, the same for its momentum P, therefore the shorter its wavelength  , and the greater will then be the resolution R.

R 1/ = P/h   (de Broglie, where h = Planck’ s constant)
P – (2mE)1/2 E  V (m = electronic mass)
 
3. The higher the temperature T, the faster the strain rate E, and the shorter will be the rupture time, under the assumption that failure occurs at the same value of strain. Arrhenius law for cheep: 
E  exp (- /kT), where  is the activation energy and k Boltzmann’s constant.
4. The ceramic, which is more brittle than the alloy, contains cracks that tend to open and propagate under tension (but not under compression.)

Griffith crack model: m 2 (c/ )1/2.

Where m is the maximum stress at the top of a crack, of length c and tip radius  , as a result of an externally applied stress  .
 
5. Brilliance is proportional to the function of light reflected at the air-diamond interface under specular reflection condition, R, and in turn R n, the refractive index of diamond.
Fresuel’s law for (near) normal incidence: 


 
6. The concentration of free electrons/holes (ne or np) in intrinsic Si rises with temperature. The same, therefore, for conductivity 6 (under the assumption of uniform electron/hole mobility, which holds approximately). Fermi function f(E) = [exp(E-Ef)/kT+1]-1, where E is electron/hole energy, Ef the Fermic level (essentially independent of temperature), k the Boltzmann constant, and T the temperature. Multiplies into the density of state to give the concentration of free electrons/holes, 
ne = np.
7. i) The slip planes are:
(2 marks)

Linear density of atoms along the  direction is (a), where a = lattice constant;    (2marks)
Trial and error show that this density is the highest among all lattice directions within the {111} planes. Therefore slip systems are .                 (4 marks)

ii) Angle between [100] and slip direction :

= 45o;                  (3 marks)

angle between [100] and normal to slip plane (111), i.e. between [100] and [111]:
= cos-1 ()     (3 marks)

so, resolved shear stress 
                                   =  *cos* cos
                                     = 1MPa * 
                                      0.41MPa                                (3 marks)

iii)  max = (1MPa) / cos/ cos

                = MPa

                 2.45 MPa                     (3 marks)
8. a)    T1 = 1010oC = 1283K

       T2 = 759oC = 1032K

       o =  e-Q /RT

         T = Q /Rln ( /o)

       1 = 104 2 = 108

         = 405 kJ/mol

(b)(c)     =oe +Q /RT

            3 = 1013.4P

            4 = 107.6P.

so,        T3 = 817K (544oC)

            T4 = 1053K (780oC)

Tempering : See Shackelford p.330 
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