Chap11 soln

Physical Chemistry

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11 Quantum theory: introduction and principles Solutions to exercises Discussion questions E11.1(b) A successful theory of black-body radiation must be able to explain the energy density distribution of the radiation as a function of wavelength, in particular, the observed drop to zero as λ 0. Classical theory predicts the opposite. However, if we assume, as did Planck, that the energy of the oscillators that constitute electromagnetic radiation are quantized according to the relation E = nhν = nhc/λ , we see that at short wavelengths the energy of the oscillators is very large. This energy is too large for the walls to supply it, so the short-wavelength oscillators remain unexcited. The effect of quantization is to reduce the contribution to the total energy emitted by the black-body from the high-energy short-wavelength oscillators, for they cannot be sufficiently excited with the energy available. E11.2(b) In quantum mechanics all dynamical properties of a physical system have associated with them a corresponding operator. The system itself is described by a wavefunction. The observable properties of the system can be obtained in one of two ways from the wavefunction depending upon whether or not the wavefunction is an eigenfunction of the operator. When the function representing the state of the system is an eigenfunction of the operator , we solve the eigenvalue equation (eqn 11.30) = ω in order to obtain the observable values, ω , of the dynamical properties. When the function is not an eigenfunction of , we can only find the average or expectation value of dynamical properties by performing the integration shown in eqn 11.39 = d τ. E11.3(b) No answer. Numerical exercises E11.4(b) The power is equal to the excitance M times the emitting area P = MA = σT 4 ( 2 πrl) = ( 5 . 67 × 10 8 W m 2 K 4 ) × ( 3300 K ) 4 × ( 2 π) × ( 0 . 12 × 10 3 m ) × ( 5 . 0 × 10 2 m ) = 2 . 5 × 10 2 W Comment . This could be a 250 W incandescent light bulb. E11.5(b) Wien’s displacement law is T λ max = c 2 / 5 so λ max = c 2 5 T = 1 . 44 × 10 2 m K 5 ( 2500 K ) = 1 . 15 × 10 6 m = 1 . 15 µ m E11.6(b) The de Broglie relation is λ = h p = h mv so v = h = 6 . 626 × 10 34 J s ( 1 . 675 × 10 27 kg ) × ( 3 . 0 × 10 2 m ) v = 1 . 3 × 10 5 m s 1
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172 INSTRUCTOR S MANUAL E11.7(b) The de Broglie relation is λ = h p = h mv so v = h = 6 . 626 × 10 34 J s ( 9 . 11 × 10 31 kg ) × ( 0 . 45 × 10 9 m ) v = 1 . 6 × 10 6 m s 1 E11.8(b) The momentum of a photon is p = h λ = 6 . 626 × 10 34 J s 350 × 10 9 m = 1 . 89 × 10 27 kg m s 1 The momentum of a particle is p = mv so v = p m = 1 . 89 × 10 27 kg m s 1 2 ( 1 . 0078 × 10 3 kg mol 1 / 6 . 022 × 10 23 mol 1 ) v = 0 . 565 m s 1 E11.9(b) The energy of the photon is equal to the ionization energy plus the kinetic energy of the ejected electron E photon = E ionize + E electron so hc λ = E ionize + 1 2 mv 2 and λ = hc E ionize + 1 2 mv 2 = ( 6 . 626 × 10 34 J s ) × ( 2 . 998 × 10 8 m s 1 ) 5 . 12 × 10 18 J + 1 2 ( 9 . 11 × 10 31 kg ) × ( 345 × 10 3 m s 1 ) 2 = 3 . 48 × 10 8 m = 38 . 4 nm E11.10(b) The uncertainty principle is p x 1 2 ¯ h so the minimum uncertainty in position is x = ¯ h 2 p =
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