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Chapter 12 Guided Review answers. 1. E ph = hf = K + E B 3.6 eV = 5.76 x 10 -19 J. f = E ph /h = 0.87 x 10 15 Hz, λ = c/f = 345 nm. 2. (a) E 3 = -13.6/9 = -1.51 eV, K = 1.51 eV, U = -2K = -3.02 eV. (b) E 5 – E 3 = 0.966 eV = 1.55 x 10 -19 J, f = 0.234 x 10 14 Hz, λ = 1280 nm 3. E’ = 0.20 MeV, E e = 0.80 MeV. 4. (a) 1 MeV – 2.1 eV = 1 MeV (b) E e max = 0.80 Mev = most probable energy, E min = 0. 5. (a) The positron comes to rest near an electron and the two annihilate. Two photons appear instead, each with an energy of 0.51 MeV. (b) The two photons are emitted in opposite directions. They have sufficient energy for most of them to leave the body and be detected outside. 6. E max = 20keV, E min = 0, f max = 4.83 x 10 18 Hz, f min = 0, λ min = 0.062 nm, λ max = . 7. ħ 2 π 2 /2mL 2 = 22.8 MeV 8. (a) a = 2m(W – E)/ ħ (b) a = 9.8 x 10 14 m -1 (c) e -ax = e -.98 = 2.66 9. (a) The alpha particle stays in the nucleus with a sinusoidal wave function. (b) There is a finite chance to find the alpha particle outside. (It tunnels to the

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Unformatted text preview: outside.) The nucleus decays by α-decay. (c) The wave function is sinusoidal both inside and outside the nucleus. The α-particle is not bound to the nucleus. 10. (a) – ħ 2 /2m d 2 Ψ /dx 2 = E Ψ (b) Both are sinusoidal. (c) Only for No. 7 is Ψ zero at x = 0 and at x = L. 11. E = ħ 2 π 2 /2mL 2 = p 2 /2m or p = ħπ /L. pL = πħ . Δ p Δ x = πħ . 12. L: orbital angular momentum, S: spin angular momentum, ℓ : orbital quantum number, L = ħ√ℓ ( ℓ + 1) s: spin quantum number, S = ħ√ s(s + 1) 13. The wave functions are different, and therefore also the electron probability distributions. 14. Z = 3, n = 1, ℓ = 0, m ℓ = 0, m s = ½, - ½; n = 2, ℓ = 0, m ℓ = 0, m s = ½, or - ½. 15. (a) The state for n = 4, ℓ = 0 is lower than that for n = 3, ℓ = 2. (b) n = 3, ℓ = 2. (These levels are filled after those with n = 4, ℓ = 0....
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