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Neutron Conversion Factors E (meV) = 81.80  -2 (Å -2 ) E (meV) = 2.072 k 2 (Å -2 ) E (meV) = 5.227  10 6  -2 (m 2 /  sec 2 ) T (K) = 11.604 E (meV)

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Presentation on theme: "Neutron Conversion Factors E (meV) = 81.80  -2 (Å -2 ) E (meV) = 2.072 k 2 (Å -2 ) E (meV) = 5.227  10 6  -2 (m 2 /  sec 2 ) T (K) = 11.604 E (meV)"— Presentation transcript:

1 Neutron Conversion Factors E (meV) = 81.80  -2 (Å -2 ) E (meV) = 2.072 k 2 (Å -2 ) E (meV) = 5.227  10 6  -2 (m 2 /  sec 2 ) T (K) = 11.604 E (meV) where  is the time of flight, or inverse velocity (  = 1/v).

2 Inelastic Scattering Processes Conservation of energy Conservation of “momentum” E i, k i E f, k f 

3 Reciprocal Space Construction The scattering triangle

4 4 Ewald Spheres kfkf kiki Q k i = 2π / λ i k f = 2π / λ f Q = k i - k f

5 From the scattering triangle, we can see that from which it follows that and so putting E f =E i -  we get This equation gives us the locus of (Q,  ) for a given scattering angle . (N.B. we can write ℏ 2 /2m=2.072 for E(meV) and Q in Å -1 ). Q -k f kiki  Kinematic Range

6 Locus of Neutrons in (Q,  ) Space  = 118°  = 2.5°  = E i Q → 2k i Q -k f kiki 

7 Scattering in ZrH 2

8 Can also be expressed in terms of the components of Q parallel and perpendicular to the incident wavevector k i : Hence it follows that and Q Q  Q kiki -k f  Q  = k f sin (  ) Q = k i – k f cos (  ) This results in the surface of a paraboloid, with the apex in (Q //, Q ,  )-space at the point (k i, 0, E i ). Kinematics in a single crystal

9 Kinematics in a Single Crystal (contd) kiki -k f Q In a single crystal experiment, we need to superimpose the scattering triangle on the reciprocal lattice. Locus of constant w is a Q-circle of radius k f centered on Q = k i [h,0,0] [0,k,0]

10 La 0.7 Pb 0.3 MnO 3 - CMR Ferromagnet QQ Q ||  kIkI kFkF Q

11 MAPS spectrometer Position sensitive detector array Specification : 20meV< E I < 2000meV l mod-chop = 10m l sam-det = 6m low angle bank: 3°-20° high angle bank: → 60° Δhω/E I = 1- 5% (FWHH) ~ 50% more flux or ~ 25% better resln. 40,000 detector elements 2500 time channels → 10 8 pixels ≡ 0.4GB datasets Background chopper Monochromating chopper Sample position cf SEQUOIA and ARCS at SNS

12 Spin Waves in Cobalt H = -J  S i.S j 12SJ = 199±7 meV  = 69 ±12 meV

13 k  c

14 Stephen Nagler (ORNL) Bella Lake(Oxford) Alan Tennant (St. Andrews) Radu Coldea(ISIS/ORNL) Direct observation of the continuum KCuF 3 Excitations (again)

15 Current Analysis Software GUI interface run info. sample 2D & 1D cuts 2D slices in (Q,  ) 1D cuts in (Q,  ) MSLICE - Radu Coldea (ISIS/ORNL)


Download ppt "Neutron Conversion Factors E (meV) = 81.80  -2 (Å -2 ) E (meV) = 2.072 k 2 (Å -2 ) E (meV) = 5.227  10 6  -2 (m 2 /  sec 2 ) T (K) = 11.604 E (meV)"

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