The Boltzmann Magetization\(M_0 = \frac{N {\gamma}^2 \hbar^2 B_0}{4 k T}\), then after elimination the units is \(J/T\).
The Polarization is \(P = \frac{\gamma \hbar B_0}{2kT}\), then after elimination we can get that \(P\) is a special number depends on the material, no SI units.
Question 2：
The polarization is \(P = \frac{51-49}{100} = 0.02\) .
The magnet field strength should be \(B_0 = \frac{0.02}{0.0000034} \approx 5882T\) .
The temperature should be \(T = \frac{300 \times 0.0000034}{0.02} = 0.051K\).
Question 3：
Since the Boltzmann Magetization Equation is \(M = M_0(1- e^{-\frac{t}{T_1}}) e^{-\frac{t}{T_2}}\) , so we can calculate the signal.
The signal of Tissue \(A\) : \(M_A = M_0(1- e^{-\frac{150}{300}}) e^{-\frac{12.5}{20}} = 0.21\) . The signal of Tissue \(B\) : \(M_B = M_0(1- e^{-\frac{150}{200}}) e^{-\frac{12.5}{40}} = 0.38\) .
Surely Tissue \(B\) will deliver more signal.
We have calculated that Tissue \(B\) will deliver more signal if both Tissue \(A\) and \(B\) has the same Boltzmann Magetization.
If Tissue \(A\) is \(85\%\) of Tissue \(B\), then the Tissue \(A\) signal will become lesser, so Tissue \(B\) deliver more signal.
Let function \(f(t) = M_{0A}(1- e^{-\frac{TR}{T_{1A}}})e^{-\frac{t}{T_{2A}}} - M_{0B}(1- e^{-\frac{TR}{T_{1B}}})e^{-\frac{t}{T_{2B}}}\) reprent the signal of time \(t\).
Consider the function: \(f(t) = (1 - e^{-\frac{150}{200}}) e^{-\frac{t}{40}} - (1- e^{-\frac{150}{300}}) e^{-\frac{t}{20}}\) reaches its PEAK at about \(t = 16\), so the \(TE\) should be \(TE = 32ms\).
% gdb -help print startup help, show switches *% gdb object normal debug *% gdb object core core debug (must specify core file) %% gdb object pid attach to running process % gdb use file command to load object
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