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542 15 SOLIDS
Substituting C = 4πV(2me/h2)3/2 gives
N = 8πV
3
(2me
h2
)
3/2
µ3/2
(iii) Rearranging the previous equatio to make µ the subject gives
µ = ( 3N
8πV
)
2/3 h2
2me
WithN = N/V gives
µ = (3N
8π
)
2/3 h2
2me
(iv) If each sodiumatomcontributes one electron to the solid,N is equal to the
number of sodium atoms in the solid.�e mass density ρ is expressed as
ρ = MN/NAV = MN /NA, whereM is themolar mass of sodium. Hence
N = ρNA/M
µ = (3ρNA
8πM
)
2/3 h2
2me
= [3 × (9.7 × 105 gm−3) × (6.0221 × 1023mol−1)
8π × (22.99 gmol−1)
]
2/3
× (6.6261 × 10−34 J s)2
2 × (9.1094 × 10−31 kg)
= 5.049... × 10−19 J = 3.2 eV
P15E.4 �e conductance,G, of germanium is predicted to have an Arrhenius-like tem-
perature dependence. �e given equation is rearranged to give a straight-line
plot
G = G0e−Eg/2kT hence lnG = lnG0 −
Eg
2kT
Hence a plot of lnG against (1/T) should be a straight linewith slope−Eg/2k.�e
data are plotted in Fig. 15.9.
T/K G/S (1/T)/K−1 ln (G/S)
312 0.084 7 0.003 21 −2.469
354 0.429 0 0.002 82 −0.846
420 2.860 0 0.002 38 1.051
Linear regression analysis gives the equation for the best-�t line as
ln (G/S) = (−4.270 × 103) × (1/T)/K−1 + 11.22
�e band gap is computed from the slope
Eg = −2k × (slope)
= −2×(1.3806 × 10−23 JK−1) × (−4.270 × 103 K) = 1.179 × 10−19 J
= 0.736 eV