Solution) (m)co = 28.0106 u
(m)N2 = 28.0134 u
D = 0.76 mm = 0.76×10^(-3) m
R = (D/2) = (0.76×10^(-3))/(2)
R = 0.38×10^(-3) m
From
Bqv = ((m)(v^2))/(r)
m = (Bqr)/(v)
(m/r) = (Bq/v)
As B , v and q are constant
(m/r)co = (m/r)N2
(28.0106/r) = (28.0134/(r + R))
(R+r)/(r) = (28.0134/28.0106)
(0.38×10^(-3) + r )/(r) = 1.0000999
0.38×10^(-3) + r = 1.0000999(r)
1.0000999(r) - r = 0.38×10^(-3)
0.0000999(r) = 0.38×10^(-3)
r = (0.38×10^(-3))/(0.0000999)
r = 3.8 m
A mass spectrometer is being used to monitor air pollutants. It is difficult, however, to separate...
ConstantsPeriodic Table Part A A mass spectrometer is being used to monitor air pollutants. It is difficult, however, to separate molecules with nearly equal mass such as CO (28.0106 u) and N2 (28.0134 u) How large a radius of curvature must a spectrometer have if these two molecules are to be separated at the film or detectors by 0.48 mm? Express your answer using two significant figures. Submit Request Answer Provide Feedback Next >
ConstantsPeriodic Table Part A A mass...
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