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242 7 QUANTUM THEORY
= 0.75), the width of the barrier, (L = 100 pm), and the decay parameter of the
wavefunction inside the barrier (κ)
κ = (2m(E − V0))1/2
ħ
= (2 × (1.6726 × 10−27 kg) × [(2.0 − 1.5) eV × 1.6022 × 10−19 J eV−1])1/2
1.0546 × 10−34 J s
= 1.58... × 1011 m−1
Such that κL = (1.58... × 1011 m−1) × (100 × 10−12 m) = 15.8..., and so the
transmission probability is given by
T = [1 + (eκL − e−κL)2
16ε(1 − ε)
]
−1
= [1 + (e15.8... − e−15.8...)2
16 × 0.75 × (1 − 0.75)
]
−1
= 5.7 × 10−14
Solutions to problems
P7D.2 (i) �e energy levels of a particle in a cubic box are given by [7D.13b–267]
E(nx , ny , nz) = h2n2/8mL2 where n2 = n2x + n2y + n2z , and L is the side
length of the box, such that its volume is V = L3. For a cubic box, volume
1.00 m3, L = (1.00 m3)1/3 = 1.00 m. �e mass of a N2 molecule is
28.00 mu = 4.6494 × 10−26 kg. To �nd the value of n that makes the
energy equal to the average thermal energy, set En = 3
2 kT and solve for n
n = L
√
12mkT
h
=
(1.00 m)×[12×(4.6494 × 10−26 kg)×(300 K)×(1.3806 × 10−23 JK−1)]1/2
6.6261 × 10−34 J s
= 7.25 × 1010
(ii) �e separation between neighbouring levels is
∆E = En+1 − En =
h2
8mL2
[(n + 1)2 − n2] = h
2(2n + 1)
8mL2
and so for this case
∆E = (6.6261 × 10−34 J s)2 × (2 × 7.25 × 1010 + 1)
8 × (4.6494 × 10−26 kg) × (1.00 m)2
= 1.71 × 10−31 J
(iii) �e de Broglie wavelength is given by [7A.11–244], λ = h/p, and the
kinetic energy is Ek = p2/2m. In this case Ek = 3
2 kT , hence p
2/2m = 3
2 kT
which is rearranged for the momentum to give p =
√
3mkT . Hence, the
de Broglie wavelength is
λ = h√
3mkT
= 6.6261 × 10−34 J s
[3×(4.6494 × 10−26 kg)×(1.3806 × 10−23 JK−1)×(300 K)]1/2
= 27.6 pm