Consider the Bainbridge mass spectrometer shown in Fig. 27.24 of the textbook (p

Question

Consider the Bainbridge mass spectrometer shown in Fig. 27.24 of the textbook (page 930).

(a) In terms of the electric and magnetic forces acting on the particles while they are in the velocity selector, prove that for a magnetic field B pointing out of the page, the electric field E has to be from left to right as shown in the figure.

(b) What is the speed v of the particles when they come out of the velocity selector?

(c) Show that the radius R of the semicircular path in the magnetic field B' is given by R = mv/qB.

(d) The radius of the semicircle for the gold ions is R2 = 20.0 cm. The mass of a gold ion is m2 = 3.27 × 1025 kg. What is the magnitude of B?

(e) The radius of the semicircle for the molybdenum ions is R1 = 9.905 cm. What is the mass of a
molybdenum ion?

Explanation / Answer

The electric field E = 1.789*10^4 V/m The magnetic field B = 1.0T The charge q =1.6*10^-19C the mass of the gold atom m = 3.27*10^-25 kg (a) From the given figure             The velocity of the atom is towards downwards and the electric force is acting towards right then the electric field should be left to right and the magnetic field is towards left to make net force is zero. Then from right hand rule force is  left and the velocity is to downwards then the magnetic field should be out of the plane of the paper. (b) When the net force is zero           then Fnet = FE + FB = 0               qE = vqB since the velocity and magnetic field are perpendicualr to each other then sin = 1         then the speed             v = E/B =  1.789*10^4 /1 = 1.789*10^4 m/s (c) When the atom moves in a circular path of radius R          then the centripetal force is equal to the magnetic force                   mv^2/R = qvB therefore the radius                 R = mv / qB (d) If R = 0.2 m   therefore the magnitude of the magnetic field                B = mv/qR                     = (3.27*10^-25kg)(1.789*10^4) / (1.6*10^-19C)(0.2)                     = 0.18 T (e) When R = 9.905*10^-2 m and B = 0.18T            therefore the mass of the molybdenum                m = qBR /v                   = (1.6*10^-19C)(9.905*10^-2)(0.18) / (1.789*10^4)                   = 1.59*10^-25 kg

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