The velocity filter of a mass spectrometer employs an electric field strength of `1.0 xx 10^(2)` V/m and a perpendicular magnetic induction of `2.0 xx 10^(-4)T.` The induction of the deflecting uniform magnolie Deld which is perpendicular to the beam is `9.0 xx 10^(-2) T.` Ions with similar arges and with mass numbers 20 and 22 pass through the filter and make a `180^@` turn in the deflecting field (Fig. 28.12). What is the distance between the points `S_1 and S_2?`

Answer

2.3mm

Explanation

The distance of interest is equal to the difference in the diameters of the ionic orbits `A= 2(R_(2)-R_(1)).` The speeds of both ions are identical and are determined by the conditions of their passage through the velocity filter: u= E/B. The radius of the orbit is `R=mu//qB_(0)`. Therefore the difference sought is `triangle=(2 E)/(qBB_(0)) (m_(2)-m_(1))`
We obtain the difference sought by assurning the ion to be singly ionized, ie. by putting: q= e.