Linked scan at constant B(1-(E/E0))1/2/E: Difference between revisions
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{{Def|Linked Scan at Constant B[1-(E/E<sub>0</sub>)]<sup>1/2</sup>/E | {{Def|Linked Scan at Constant B[1-(E/E<sub>0</sub>)]<sup>1/2</sup>/E | ||
|A linked scan at constant B[1-(E/E<sub>0</sub>)]<sup>1/2</sup>/E | |A linked scan at constant B[1-(E/E<sub>0</sub>)]<sup>1/2</sup>/E is performed on a [[sector instrument]] that incorporates at least one [[electric sector]] plus one [[magnetic sector]] placed in either order. The [[accelerating voltage]] is fixed while scanning the magnetic field, B, and electric field, E, simultaneously, so as to maintain the quantity B[1-(E/E<sub>0</sub>)]<sup>1/2</sup>/E at a constant value. This [[linked scan]] records a [[constant neutral mass loss (or gain) spectrum]] of dissociation or other reactions occurring in a [[field free region]] (FFR) preceding the two sectors. E<sub>0</sub> is the electric field required to transmit the singly charged analog of the desired neutral fragment. The term "B[1-(E/E<sub>0</sub>)]1/2/E linked scan" is not recommended. | ||
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Revision as of 12:26, 14 December 2005
DRAFT DEFINITION |
Linked scan at constant B(1-(E/E0))1/2/E |
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This is an unofficial draft definition presented for information and comment. |
Orange Book Entry
This can be performed with a sector instrument incorporating at least one electric sector plus one magnetic sector. The magnetic sector field strength, B, and the electric sector field strength, E, are scanned simultaneously, while holding the accelerating voltage , V, constant, so as to maintain the ratio B [1- (E /E 0)]1/2 /E constant. This constant value is equal to B3/E0, where E0 and B3 are respectively the electric and magnetic sector fields required to transmit M3+ ions in the main ion beam; m3 represents the mass (m1-m2) of the selected neutral fragment M3 whose precursor spectrum is required. The fragmentation reactions thus observed occur in a field-free region traversed before the two sectors scanned in this way. The term 'B [1- (E /E 0)]1/2/E linked scan????????????????????????? is not recommended.