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238 7 QUANTUM THEORY
E7D.4(b) �e wavefunctions are ψ1(x) = (2/L)1/2 sin(πx/L) and ψ3(x) = (2/L)1/2 ×
sin(3πx/L). �ey are orthogonal if ∫ ψ∗1ψ2 dτ = 0. In this case the integral is
taken from x = 0 to x = L as outside this range the wavefunctions are zero.�e
required integral is of the form of Integral T.5, with A = π/L, B = 3π/L, and
a = L
2
L ∫
L
0
sin(πx
L
) sin(3πx
L
)dx = 2
L
[ sin(−2πL/L)
−4π/L
− sin(4πL/L)
8π/L
]
= 2
L
[ sin(−2π)
−4π/L
− sin(4π)
8π/L
] = 0
where sin nπ = 0 for integer n is used.�e two wavefunctions are orthogonal.
E7D.5(b) �e particle in a boxwavefunctionwith quantumnumber n is given byψn(x) =
(2/L)1/2 sin(nπx/L)�e probability of �nding the electron in a small region of
space δx centred on position x is approximated as ψ2(x)δx. For this Exercise
x = 0.66L, δx = 0.02L.
For the case where n = 1
ψ1(0.66L)2 × 0.02L = [
√
2/L sin (π(0.66L)/L)]
2
× 0.02L
= (2/L) sin2(0.66π) × 0.02L = 0.031
For the case where n = 2
ψ2(0.66L)2 × 0.02L = [
√
2/L sin (2 × π(0.66L)L)]
2
× 0.02L
= (2/L) sin2(1.32π) × 0.02L = 0.029
E7D.6(b) �e wavefunction with n = 2 and is ψ2(x) =
√
2/L sin(2πx/L), which leads to
a probability density P2(x) = ∣ψ2(x)∣2 = (2/L) sin2(2πx/L). Graphs of these
functions are shown in Fig 7.4.
0.2 0.4 0.6 0.8 1.0
x/L
ψ2
ψ22
Figure 7.4
�e probability density is symmetric about x = L/2.�erefore, there is an equal
probability of observing the particle at an arbitrary position x′ and at L− x′, so
it follows that the average position of the particle must be at L/2.