Question

An accelerating voltage of 7.00 kV is used to accelerate ions into the mass analyzer tube of a magnetic sector mass spectrometer, which has a radius of curvature of 9.49 cm. If ions of m/z 531.8 reach the detector, what is the strength of the magnetic field that is applied perpendicular to the length of the mass analyzer tube?

Answer 1

Use the following relation to find out the strength of the magnetic field applied:

m/z = 4.825*10^-5*B²r²/V where m/z is the mass per unit charge of the ion recorded; B = magnetic field strength in Gauss, r = radius of curvature of the path of the ion in cm and V = voltage applied in Volts.

Given V = 7.00 kV = 7.00*1000 V, plug in values and obtain

531.8 = 4.825*10^-5*B²*(9.49)²/(7.00*1000)

===> B² = (531.8*7.00*1000)/(4.825*10^-5).(9.49)² = 8.567*10⁸

===> B = 2.927*10⁴

The strength of the magnetic field is 2.927*10⁴ gauss = (2.927*10⁴ gauss)*(1 T/10⁴ gauss) = 2.927 T ≈ 2.93 T (ans).