In other be able to justify a sample that can be used as contrast agent in MRI, so some requirement must be taken into account, such as magnetic properties, optical properties, and relaxation time T1 and T2. I don't understand yet how to characterize sample relaxation time T1 and T2?
Dear Sir,
you can use the permittivity measurements. With the complex permittivity and real permittivity over the frequency. Use this eps ´and eps ´´ for a cole cole plot diagramm.
After this diagram you can use the tangent line and the angular is proportional to the T1 and T2.
All the best
F. Gräbner
I am not sure that I really understand your question: I assume you are talking about magnetic nanoparticles that are to be used as magnetic resonance (MR) contrast agents?
MR contrast agents (CAs) have the property that they shorten the relaxation time constants T1 and T2 of e.g. water. I.e. pure water (or biological tissue) without any contrast agent has typical values of T1,0 and T2,0, and after adding some contrast agent (resulting in a CA concentration c), T1 and T2 are reduced depending on c and the so-called relaxivities r1 and r2:
1 / T1(c) = 1 / T1,0 + c × r1
1 / T2(c) = 1 / T2,0 + c × r2
So, CAs are characterized by their relaxivities r1 and r2, which can be determined by T1 and T2 measurements. For the latter, please see the questions and answers referenced below.
(Optical properties are not relevant for MR contrast agents.)
https://www.researchgate.net/post/How_can_I_measure_relaxivities_of_mri_constrast_agents
Dear all
I really understand now about the contrast agent, so the explanations are so clear. Thank you very much.
Best wishes
How can I calculate the concentration of CA if I know the T1 of the tissue , the T1 after applying CA and relaxitivity of CA?
Mohamed A. ElSharawy - You can calculate the CA concentration by solving the equations given above, which yields:
cCA = (1 / T1,withCA – 1 / T1,beforeCA) / r1.
Example:
T1,beforeCA = 500 ms
T1,withCA = 200 ms
r1 = 5 L/mmol/s
cCA = (1 / 0.2 s – 1 / 0.5 s) / 5 L/mmol/s = (5 s-1 – 2 s-1) / 5 L/mmol/s = 3/5 mmol/L = 0.6 mmol/L
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