Question

In a mass spectrometer, a singly charged ion having a particular velocity is selected by using a magnetic field of 7.0×10−2 T perpendicular to an electric field of 1.0×103 V/m . A magnetic field of this same magnitude is then used to deflect the ion, which moves in a circular path with a radius of 1.3 cm .

What is the mass of the ion?

Express your answer using two significant figures.

Answer 1

In a mass spectrometer, the velocity of a ion is selected by applying a magnetic field (B) at right angles to the electrostatic field, so the electrostatic and electromagnetic forces act in opposite directions to each other. The velocity of the ion is determined by balancing these two opposite forces.

  

$\text{Electromagnetic force } (B_sqv) = \text{Electrostatic force } {(qE_s)} \\ \Rightarrow \text{velocity of the ion } (v) = \frac{E_s}{B_s} = \frac{E}{B}$

where the subscript denotes the velocity selector field.

After the velocity is selected, the charge (q) moves into another magnetic field (B same magnitude) with direction perpendicular to the field, it will follow a circular path. The magnetic force, being perpendicular to the velocity, provides the centripetal force and we have,

$\frac{mv^2}{r} = q vB \Rightarrow m = \frac{qrB}{v} = \frac{qrB^2}{E}$

Here the ion is singly charged i.e. q = 1.6*10^-19 C, r =0.013 m, B = 7*10^-2 T, E = 1000 V/m

Just put these values in the expression of mass m and get,

$m = \frac{1.6\times 10^{-19}\times 0.013\times (0.07)^2}{1000}= 1.02 \times 10^{-26}\hspace{2mm} kg$