# HW9.pdf - 194 CHAPTER 12 12-15(a THE SOLID STATE Eg = 1.14...

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194 CHAPTER 12 THE SOLID STATE 12-15 (a) = 1.14 eV g E for Si ( ) ( ) = = × = × = × 19 19 14 1.14 eV 1.14 eV 1.6 10 J eV 1.82 10 J 2.75 10 Hz hf f (b) λ = c f ; λ × = = = × × 8 6 14 3 10 m s 1.09 10 m 2.75 10 Hz c f λ = 1 090 nm (in the infrared region) 12-16 λ × = = = = 3 min max 1.240 10 eV nm 0.670 eV 1 850 nm g hc E hf 12-17 (a) R < U II I III U x = 0 x = a Potential x ( ) [ ] ( ) ψ ψ ψ = = = + = = = = 1 2 I 1 2 II III 2 cos sin 2 Kx Kx Ae K m U E B kx C kx k mE De In region I and III the wave equation has the form ( ) ( ) ψ ψ = 2 2 2 d x K x dx with ( ) [ ] = = 1 2 2 m U E K . This equation has solutions of the form ( ) ( ) ( ) ψ ψ = = I III for 0 region I for 0 region III Kx Kx x Ae x De x In region II where ( ) = 0 U x we have ( ) ( ) ψ ψ = − 2 2 2 d x k x dx with [ ] = = 1 2 2 mE k . This equation has trigonometric solutions ( ) ψ = + II cos sin 0 x B kx C kx x a with ( ) = = 1 2 2 mE k . The wave function and its slope are continuous everywhere, and in particular at the well edges = 0 x and = x a . Thus, we must require

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203 13 Nuclear Structure 13-1 = 1 3 0 R R A where = 0 1.2 fm R ; (a) = 4 A so ( )( ) = = 1 3 He 1.2 4 fm 1.9 fm R (b) = 238 A so ( )( ) = = 1 3 U 1.2 238 fm 7.44 fm R (c) = = U He 7.44 fm 3.92 1.9 fm R R 13-2 ( )( ) ρ = = × = × 14 3 3 15 2.3 10 g cm 10 cm 2.3 10 kg m V !!! 13-3 ρ ρ = NUC NUC NUC ATOMIC ATOMIC ATOMIC M V M V and approximately; = NUC ATOMIC M M . Therefore ρ ρ = 3 NUC 0 ATOMIC r R where = 0 0.529 r Å = × 11 5.29 10 m and = × 15 1.2 10 m R (Equation 13.1 where = 1 A ). So that ρ ρ × = = × × 3 11 13 NUC 15 ATOMIC 5.29 10 m 8.57 10 1.2 10 m . 13-4 (a) ( ) ( ) ( ) µ × = = = × 27 34 2 1.913 5 5.05 10 J T 1 T 2 29.1 MHz 6.63 10 J s n B f h (b) ( ) ( ) ( ) × = = × 27 34 2 2.792 8 5.05 10 J T 1 T 42.5 MHz 6.63 10 J s p f (c) In the earth’s magnetic field ( ) ( )( ) × × = = × 27 6 34 2 2.792 8 5.05 10 J T 50 10 T 2.13 kHz 6.63 10 J s p f .
204 CHAPTER 13 NUCLEAR STRUCTURE 13-5 (a) The initial kinetic energy of the alpha particle must equal the electrostatic potential energy of the two particle system at the distance of closest approach; α = = min kqQ K U r and ( ) ( ) ( ) ( ) α × × = = = × × 2 9 2 2 19 13 min 13 9 10 N m C 2 79 1.6 10 C 4.55 10 m 0.5 MeV 1.6 10 J MeV kqQ r K . (b) Note that α = = 2 min 1 2 kqQ K mv r , so ( ) ( ) ( ) ( )( ) × × = = = × × × 1 2 2 9 2 2 19 1 2 6 27 13 min 2 9 10 N m C 2 79 1.6 10 C 2 6.03 10 m s 4 1.67 10 kg 3 10 m kqQ v mr 13-6 (a) = 1 3 0 R R A , = × 15 0 1.2 10 m R , = 4 10 m R ( ) ( )( ) = = = × = × × = = × × = × 3 3 4 3 18 56 -15 0 27 56 29 n 10 m 8.33 10 5.8 10 1.2 10 m 1.67 10 kg 5.8 10 9.7 10 kg R A R M m A (about 1/2 the mass of the sun) (b) = 2 GMm mg R so = 2 GM g R ( )( ) ( ) × × = = × 11 2 2 29 11 2 2 4 6.67 10 N m kg 9.7 10 kg 6.5 10 m s 10 m g (c) ω = 2 1 2 K I ( ) ω π = × 30 2 rad s ; and ( )( ) = = × = × 2 2 29 4 37 2 2 2 9.7 10 kg 10 m 3.88 10 kg m 5 5 I MR so ( ) ( ) ω π = = × × = × 2 2 37 2 41 1 1 3.88 10 kg m 30 2 rad s 6.9 10 J 2 2 K I . 13-7 µ = − E B so the energies are µ = + 1 E B and µ = − 2 E B . µ µ = 2.792 8 n and µ = × 27 5.05 10 J T n µ = = × × × × = × = × 27 25 6 2 2 2.792 8 5.05 10 J T 12.5 T 3.53 10 J 2.2 10 eV E B 13-8 Note: Let the proton’s magnetic moment be µ µ = nuc 2.8 p . The neutron’s magnetic moment is µ µ = − nuc 1.9 n .

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MODERN PHYSICS 205 2.8 µ nuc –2.8 µ nuc B = 0 B > 0 0.9 µ nuc = µ p + µ n B = 0 B > 0 –0.9 µ nuc = – µ p µ n 4.7 µ nuc = µ p µ n –4.7 µ nuc = –
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