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250 7 QUANTUM THEORY
E7E.5(b) �e zero-point energy of a harmonic oscillator is given by [7E.5–274], E0 =
1
2 ħω, where the frequency ω is given by [7E.3–274], ω = (kf/µ)1/2.�e e�ec-
tive mass µ of a diatomic AB is given by [7E.6–274], µ = (mAmB)/(mA +mB).
In the case of a homonuclear diatomic A2 this reduces to µ = mA/2. For this
system,
E0 = 1
2 × (1.0546 × 10−34 J s)
× [(2293.8 Nm−1)/( 12 × 14.0031 × 1.6605 × 10
−27 kg)]1/2
= 2.3422 × 10−20 J
E7E.6(b) �e energy levels of a harmonic oscillator are [7E.3–274], Eυ = (υ+ 12 )ħω, with
ω = (kf/m)1/2 and υ = 0, 1, 2....�e energy of the state with υ = 1 is E1 = 3
2 ħω.
(i) For the system with kf = 1000 Nm−1 the energy of the state with υ = 1 is
E1 = 3
2 × (1.0546 × 10−34 J s) × [(1000 Nm−1)/(1 × 1.6605 × 10−27 kg)]1/2
= 1.22... × 10−19 J = 1.23 × 10−19 J
�e classical turning points of this state occur when E1 = 1
2 kfx
2
tp. Solving
this for xtp leads to xtp = ±
√
2E1/kf , giving a separation of
2
√
2E1/kf = 2
√
2 × (1.22... × 10−19 J)/(1000 Nm−1) = 31.3... pm
As a percentage of the typical bond length, 110 pm, this is
(31.3... pm)/(110 pm) × 100% = 28.5%
(ii) For the system with kf = 100 Nm−1 the energy of the state with υ = 1 is
E1 = 3
2 × (1.0546 × 10−34 J s) × [(100 Nm−1/(1 × 1.6605 × 10−27 kg)]1/2
= 3.88... × 10−20 J = 3.88 × 10−20 J
�e classical turning points of this state occur when E1 = 1
2 kfx
2
tp. Solving
this for xtp leads to xtp = ±
√
2E1/kf , giving a separation of
2
√
2E1/kf = 2
√
2 × (3.88... × 10−20 J)/(100 Nm−1) = 55.7... pm
As a percentage of the typical bond length, 110 pm, this is
(55.7... pm)/(110 pm) × 100% = 50.7%
E7E.7(b) �e wavefunctions are depicted in Fig. 7E.6 on page 276. �e general form of
the harmonic oscillator wavefunctions is ψυ = NυHυ(y)e−y
2/2, where Nυ is
a normalization constant and Hυ(y) is the Hermite polynomial of order υ in
the reduced position variable y. Nodes occur when the wavefunction passes