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Hand In 10 Solutions (07)

# Hand In 10 Solutions (07) - Solutions for Hand-In Set#12...

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Solutions for Hand-In Set #12 PHYS 212 – Spring 2007 A81 - Flipping inside atoms The energy of the photon is equal to the energy dif- ference between the two energy levels of the antiparallel spin state and the parallel spin state. The energy of a photon is E photon = hf = h c λ , The energy difference between the spin states is Δ E = 2 μ e B. Equating these expressions gives hc λ = 2 μ e B, or B = hc 2 μ e λ = 6 . 63 × 10 - 34 J · s × 3 × 10 8 m/s 2 × 9 . 24 × 10 - 24 J/T × 0 . 21 m = 0 . 051 T . A82 — MRI The energy of the photon is equal to the energy dif- ference between the two energy levels of the spin states. The energy of a photon is E photon = hf, and the energy difference between the spin states is Δ E = 2 μ p B. Equating these expressions gives hf = 2 μ p B. In this problem the field B is a function of position: B = 0 . 5 + 0 . 6 x . We can insert this expression into the relationsip above to relate position and frequency: hf = 2 μ p (0 . 5 + 0 . 6 x ) , and solving for position gives x = 1 0 . 6 hf 2 μ p - 0 . 5 = 1 0 . 6 6 . 63 × 10 - 34 J · s × 30 × 10 6 Hz 2 × 1 . 41 × 10 - 26 J/T - 0 . 5 = 0 . 34 m (b) If we want to probe at x = 0 . 50 m, then we can use the relationship between frequency and position from above to write f = 2 μ p h (0 . 5 + 0 . 6 x ) = 2 × 1 . 41 × 10 - 26 J/T 6 . 63 × 10 - 34 J · s (0 . 5 + 0 . 6 × 0 . 5) = 34 . 0 MHz . (c) The energy of photons for this result is E photon = hf = 6 . 63 × 10 - 34 J · s × 34 × 10 6 Hz = 2 . 25 × 10 - 26 J = 1 . 41 × 10 - 7 eV . A83 — Particle Decay (a) One unit of energy is left. This represents kinetic energy to be shared between the two particles that result from the decay. (b) You can’t do this — you don’t have enough energy. A101 — Precessing Spins II (a) From Chapter 2 in the supplementary reading we know | ψ (0) i = | + y i = r 1 2 | + z i + i r 1 2 |- z i . (b) E + = - μB 0 , E - = + μB 0 . (c) | ψ ( t ) i = r 1 2 e - iE + t/ ¯ h | + z i + i r 1 2 e - iE - t/ ¯ h |- z i = r 1 2 e iμB 0 t/ ¯ h | + z i + i r 1 2 e - iμB 0 t/ ¯ h |- z i = r 1 2 e iωt/ 2 | + z i + i r 1 2 e - iωt/ 2 |- z i .

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(d) Prob + y = |h + y | ψ ( t ) i| 2 = r 1 2 h + z | - i r 1 2 h- z | !
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