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Answers to Chapter 14 Problems

14.5.1 (a) 10¹⁵ (b) ≈12.7 days

14.5.1 Solution

14.5.2 (i) 1.82 eV (ii) 658 nm

14.5.2 Solution

14.5.3 0.44 eV; 1.24 × 10¹³

14.5.3 Solution

14.5.4 1.94 eV; 303 nm

14.5.4 Solution

14.6.1 From this data the Planck constant is estimated to be 6.56 × 10^–34 J s

14.6.1 Solution

14.7.1

14.7.1 Solution

14.7.2 4.86 × 10^–12 m

14.7.2 Solution

14.7.3 1.37 × 10^-21kg m s^–1; 4.84 × 10^{–13&�Բ��;}m; 6.20 × 10^{–20&�Բ��;}Hz

14.7.3 Solution

14.7.4 (a) 2.13 × 10^–12 m (b) 0.082

14.7.4 Solution

14.7.5 (a) 1.29 × 10^–11 m (b) 58.7 keV

14.7.5 Solution

14.7.6 (a) 2.43 × 10^–7 m (b) 458 eV

14.7.6 Solution

14.8.1 (a) photon: 8.29 × 10^–7 m; electron: 1.00 × 10^–9 m (b) photon: 8.29 × 10^–16 m; electron: 8.29 × 10^–16 m

14.8.1 Solution

14.8.2 3.4 nm; 0.074 nm (a) 1.0 × 10¹⁰ Pa (b) 0.029 K

14.8.2 Solution

14.8.3 outside: 3.9 × 10^–10 m; inside: 2.6 ×10^–10 m; 28.1°

14.8.3 Solution

14.9.1 0.51 MeV; v_g = 0.77c; v_w = 1.30c

14.9.1 Solution

14.9.2 939 MeV; (a) 3.06 × 10^–19 kg m s^–1 (b) 2.17 × 10^–15 m (c) 0.52c (d) 1.92c

14.9.2 Solution

14.10.1 158 m s^–1; 790 m

14.10.1 Solution

14.10.2 3.4 eV

14.10.2 Solution

14.10.3 (a) 0.40 m (b) 0.4 mm (c) 0.4 μm

14.10.3 Solution

14.10.4 D = \( \sqrt{{2 \hbar \over m }{\sqrt{2H \over g}}}\); 2.6 × 10^–16 m

14.10.4 Solution

14.11.1 <x> = ½ ; 0.049

14.11.1 Solution

14.11.2 \( \sqrt{2} \); (a) 0.36 (b) 0.674 (c) 0.667 (d) 0.5

14.11.2 Solution

14.13.1 0.33 m; It would not be feasible to play tennis in a universe in which the behaviour of macroscopic objects is governed by quantum, rather than classical, mechanics.

14.13.1 Solution

14.13.2

14.13.2 Solution

14.15.1 2.51 × 10^–11 m

14.15.1 Solution

14.15.2 (a) 0.006 (b) 0.20. The results indicate that the particle is more likely to be in the middle of the well than at an edge. This contrasts with the classical picture in which the probability is expected to be constant throughout the well. The results are unlike the classical expectation which is that the particle spends equal times in equal lengths of the well.

14.15.2 Solution

14.15.3 (a) For n = 1 where R_n = 3. (b) As n → ∞, R_n → 0 which for practical purposes is the same as the classical case.

14.15.3 Solution

14.15.4 \( E_n = {\pi^2 \hbar^2 n^2 \over 2ma^2} \)

14.15.4 Solution

14.15.5 n = 2 and n = 4; 7.09 × 10⁻¹⁰ m; E₂ = 3.0 eV and E₃ = 6.75 eV

14.15.5 Solution

14.15.6 0.56

14.15.6 Solution

14.15.7 1.38 × 10⁻³⁰ m s⁻¹; 1.45 × 10³⁰

14.15.7 Solution

14.16.1 0.901A; 1.290 rad

14.16.1 Solution

14.16.2 (a) T = 0.63 and R = 0.37 (b) T = 0.93 and R = 0.07

14.16.2 Solution

14.16.3 (a) In both cases (i) and (ii), R = 1 (b) (i) 4.9 × 10⁻¹¹ m (ii) 1.2 × 10⁻¹⁰ m

14.16.3 Solution

14.16.4 0.34

14.16.4 Solution

14.18.1 0.368

14.18.1 Solution

Understanding Physics

Mansfield and O'Sullivan, Understanding Physics, 3rd ed., John Wiley & Sons, Chichester (2020),

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