1. The problem statement, all variables and given/known data The diagram shows a mass spectrometer used for measuring the masses of isotopes. It consists of an ion generator and accelerator, a velocity selector and an ion separator, all in a vacuum. In one experiment, tin ions, each of which carries a charge of +1.6 * 10-19 C, are produced in the ion generator and are then accelerated by a p. d. of 20 000 V. Tin has a number of isotopes, two of which are tin-118 (118Sn) and tin-120 (120Sn). (a) (i) Assuming that an ion of tin-120 is at rest before being accelerated, show that the final speed after acceleration is 177 km s-1. Mass of nucleon = 1.7 * 10-27. (a) (ii) What will be the final speed of an ion of tin-118? (b) In practice all ions produced by the ion generator have a range of speeds. A velocity selector is used to isolate ions with a single speed. In the velocity selector the force produced by the electric field is balanced by that due to the magnetic field which is perpendicular to the plane of the paper. (b) (i) The plates producing the electric field have a separation of 2.0 cm. The potentials of the plates are marked on the diagram. What is the magnitude of the force on an ion due to this electric field in the velocity selector? (b) (ii) Write down the equation which must be satisfied if the ions are to emerge from the exit hole of the velocity selector. Define the terms in the equation. (b) (iii) What magnetic flux density is required if ions travelling with a speed of 177 km s-1 are to be selected? (c) After selection the ions are separated using a magnetic filed on its own, as shown in the diagram. (c) (i) Explain why the ions move in circular paths in this region. (c) (ii) Show that the radius of the path is directly proportional to the mass of the ion. (c) (iii) The ions are detected using the photographic plate P. Determine the distance between the points of impact on the photographic plate of the two isotopes of tin when a magnetic flux density of 0.75 T is used in the ion separator. Answers: (a) (ii) 179 m s-1, (b) (i) 3.2 * 10-15 N, (iii) 0.11 T, (c) (iii) 0.50 cm. 2. The attempt at a solution (a) (i) (1 / 2) M v2 = Q V, where Q = 1.6 * 10-19 C, V = 20 000 V, v = ? and M = ?. We find M first: Relative atomic mass Ar = Mass of atom M / (1 / 12( the mass of a 126C atom u → Ar = M / u → M = Ar * u = 120 * 1.7 * 10-27 = 2.04 * 10-25 kg. So v = √ 2 Q V / M = √ 2 * 1.6 * 10-19 * 20 000 / 2.04 * 10-25 = 177 123 m s-1 or 177 km s-1. (a) (ii) Doing the same thing, find M first: M = Ar * u = 118 * 1.7 * 10-27 = 2.006 * 10-25 kg. Plug in: v = √ 2 Q V / M = √ 2 * 1.6 * 10-19 * 20 000 / 2.006 * 10-25 = 178 617.7 m s-1. I didn't continue since it's either a typo in my book answer, or I did something wrong. I get 179 km s-1, not 179 m s-1.