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In a magnetic sector (single-focusing) mass spectrometer, it might be reasonable under some circumstances to monitor one m/z value, to then monitor a second m/z. and to repeat this pattern in a cyclic manner. Rapidly switching between two accelerating voltages while keeping all other conditions constant is called peak matching
(a) Derive a general expression that relates the ratio of the accelerating voltages to the ratio of the corresponding m/z values.
(b) Use this equation to calculate m/z of an unknown peak if m/z of the ion used as a standard, CF3+, is 69.00 and the ratio of V unknown /V standard is 0.965035.
(c) Based on your answer in part (b), and the assumption that the unknown is an organic compound that has a mass of 143, draw some conclusions about your answer in part (b), and about the compound.

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Answer:

A) (m/Z)s /  ( m/Z)u =  Vu / Vs

B)   71.5

C) the ratio of accelerating voltages decrease as the value of the unknown mass increases.

Explanation:

A) To derive the expression we have to vary one of the three variables while keeping the other two variables constant

let the variables be denoted as ; ( B , V , r )

m/z = B^2r^2e / 2v

therefore the equations can be written as

[tex]( \frac{m}{z} )_{s} = \frac{k}{v_{s} }[/tex]   ,    [tex]( \frac{m}{z} )_{u} = \frac{k}{v_{u} }[/tex]

u = unknown value , s = standard value

∴ ratio of accelerating voltage to the ratio of corresponding m/Z values

  = (m/Z)s /  ( m/Z)u =  Vu / Vs

B) find (m/Z) u

since : Vu / Vs = 0.965035

           ( m/Z) s =  69.00

Vu / vs = (m/Z)s /  (m/Z)u -------- ( 1 )

∴ (m/Z)u  =  69 / 0.965035 =  71.5

C) If the unknown mass  ( m/z ) u = 143

the ratio of the voltages  : Vu / vs = 69 / 143 = 0.4825

when compared to the answer gotten in b above it can be seen that the ratio of accelerating voltages decrease as the value of the unknown mass increases.

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