Assignment #1. Answers - Final

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Florida International University *

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4011

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Electrical Engineering

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Apr 3, 2024

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Assignment #1 Answers 1. Complete the following table for three different elements: Resonant Frequencies ω L / 2 π (MHz) Element γ / 2 π (MHz/T) B o = 1.5T B o = 4.7 T 1 H 42.58 63.87 200.1 23 Na 11.27 16.91 52.97 31 P 17.25 25.87 81.08 A total of eight items must be defined  10 points – having all items correct  8 points – missing 1-2 items (e.g., Bo=4.7 T)  5 points – missing/incorrect half of the answers  0 points – missing/incorrect less than half of the answers 2. Determine the nuclear magnetic moment for the element 15 N, using the nuclear shell model (Pauli Exclusion Principle). Deduction Protons Z = 7 (One unpaired proton on the higher energy state of the 2 nd level) 1 st level: l = 0 , 2 states 2 nd level: l = 1 , 4 states 2 nd level: l = 1 , 1 unpaired state ( i = 1 / 2 ) Neutron N = 8 ("Zero net magnetic moment due to pairing in their energy levels") 1 st level: l = 0 , 2 states 2 nd level: l = 1 , 4 states 2 nd level: l = 1 , 2 states Final calculations i = 1 / 2 I z = ±ħ / 2 = ± 0.53 Js I = √ 3 ħ / 2 = 1.73 × 6.62 × 10 − 34 Js 4 × 3.14 = 0.91 × 10 − 34 Js  10 points – deduction and final calculations correct  5 points – deduction correct, but final calculations incorrect or no deduction but inferred from text  0 points – deduction and final calculations incorrect

3. Calculate the gyromagnetic ratio ( γ / 2 π ) for the previous element and compare to the value reported on the table as shown in the class example. Deduction Unpaired Proton – 2 nd Level ( l = 1 ; p =− 1 ¿ g L = 1 ; g S = 5.586 g J = g L ± g S − g L 2 l + 1 = 1 − 5.586 − 1 2 ∗ 1 + 1 =− 0.529 m p = 1.672 × 10 − 27 kg e = 1.602 × 10 − 19 C Final calculations γ = g J e / 2 m = − 0.529 ∗ 1.602 × 10 − 19 C 2 ∗ 1.672 × 10 − 27 Kg =− 25.34 × 10 6 rad / sT γ 2 π =− 4.035 MHz T ≈ − 4.316 MHz / T → 15 N  10 points – deduction and final calculations correct  5 points – deduction correct, but final calculations incorrect or inferred from text  0 points – deduction and final calculations incorrect Note: - 2 points if you selected the + sign in the Lande factor (related to the polarity) 4. What is the magnitude of the magnetic field B 0 given a ratio of unpaired nucleons ( N + ¿ / N − ¿¿ ¿ ) in the element 13 C of 0.9999922935 at 300 K of temperature? Solution N + ¿ / N − ¿ ¿ e − γ ħB z / kT ¿¿ − ln ( 0.9999922935 ) ∗ kT γħ = B z = 4.5 T  10 points – solution correct  5 points – used formula is correct, but final solution is incorrect  0 points – solution incorrect 5. Name the element at which we need to apply a polarized magnetic field B 1 of 48.86 μT for 3 ms to obtain a longitudinal magnetization of M z = 0. Solution M z is zero with a 90 o pulse ( ∆θ = π / 2 ) γ = ∆θ B 1 τ = 1.57 48.86 μT ∗ 3 ms = 10.7 MHz / T 13 C is the valid answer  10 points – deduction and solution are correct

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 5 points – used formula is correct, solution is incorrect  0 points – deduction and solution are incorrect Note – 2 points if using 90 o instead of π⁄2