Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống
1
/ 70 trang
THÔNG TIN TÀI LIỆU
Thông tin cơ bản
Định dạng
Số trang
70
Dung lượng
0,96 MB
Nội dung
Neuroradiology MRI Graduate Entry Medicine Programme Daniel Bulte Objectives • • • • • • How does MRI work? What is resonance? Where does “the signal” come from? What are T1 and T2? Where does contrast come from? Please ask questions! Spin • Fundamental property of particles, like mass and charge • Spin comes in quanta of 1/2 (for Fermions) • Electrons, protons and neutrons all have spin 1/2 • Pairs of these subatomic particles tend to align with opposite spin and “cancel out” Nuclear Spin Some nuclei have Spin If a nucleus has an unpaired proton it will have spin and it will have a net magnetic moment or field ⇒ NMR phenomenon Common NMR Active Nuclei Isotope Spin I % abundance γ MHz/T H H 13 C 14 N 15 N 17 O 19 F 23 Na 31 P 1/2 1/2 1/2 5/2 1/2 3/2 1/2 99.985 0.015 1.108 99.63 0.37 0.037 100 100 100 42.575 6.53 10.71 3.078 4.32 5.77 40.08 11.27 17.25 γ = gyromagnetic ratio Alignment of Spins in a Magnetic Field M magnetic moment M=0 spin B0 field Energy in a Magnetic Field (Zeeman Splitting, Spin ½) mI = +½ E+1/2= −γ B0/2 mI = −½ P+1/2= 5000049 P(E) ∝ exp(−E/kT) E-1/2= +γ B0/2 P-1/2= 0.4999951 Larmor Frequency mI = −½ mI = +½ E+1/2= −γ B0/2 E-1/2= +γ B0/2 Allowed transitions ∆E = γ B0 = ω0 ω0 = γ B Larmor Frequency z B0 ω0 M0 Spins “precess” at Larmor frequency Net magnetisation M0 is static x ω = γ B0 Precession QuickTime™ and a YUV420 codec decompressor are needed to see this picture ... abundance γ MHz/T H H 13 C 14 N 15 N 17 O 19 F 23 Na 31 P 1/ 2 1/ 2 1/ 2 5/2 1/ 2 3/2 1/ 2 99.985 0. 015 1. 108 99.63 0.37 0.037 10 0 10 0 10 0 42.575 6.53 10 . 71 3.078 4.32 5.77 40.08 11 .27 17 .25 γ = gyromagnetic... Recovery Curves Short T1 (white matter) Mz Medium T1 (grey matter) Contrast Long T1 (CSF) TR T1 Weighted Image white matter grey matter CSF T1/s R1/s -1 0.7 1. 43 0.25 1. 5T SPGR, TR =14 ms, TE=5ms, flip=20º... Splitting, Spin ½) mI = +½ E +1/ 2= −γ B0/2 mI = −½ P +1/ 2= 5000049 P(E) ∝ exp(−E/kT) E -1/ 2= +γ B0/2 P -1/ 2= 0.49999 51 Larmor Frequency mI = −½ mI = +½ E +1/ 2= −γ B0/2 E -1/ 2= +γ B0/2 Allowed transitions