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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). Part A How large a radius of curvature must a spectrometer have if these two molecules are to be separated at the film or detectors by 0.42 mm ? Express your answer using two significant figures.

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IthaloAbreu

Answer:

4.2 m

Explanation:

The energy (E) of the selector region of the spectrometer is:

E = v*B (equation 1)

Where v is the velocity, and B the magnetic field.

The force with which particle forms a curved path can be calculated by:

[tex]\frac{mv^2}{r} = qB'[/tex]

Where r is the radius of the particle, m is the mass, and q the electric charge. So:

[tex]r = \frac{mv}{qB'}[/tex]

For equation 1: v = E/B, so:

[tex]r = \frac{mE}{qBB'}[/tex]

The two species will be separated by Δr = 0.42 mm

Δr = Δm*E/(q*B*B')

E/(q*B*B') = r/m

Δr = Δm*r/m

r = m*Δr/Δm

r is the large of curvature, m is the avarege mass = (28.0106 + 28.0134)/2 = 28.012:

r = (28.012*0.42)/(28.0134 - 28.0106)

r = 11.76504/(0.0028)

r = 4201.8 mm

r = 4.2 m

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