A linear accelerator (linac) accelerates charged particles in a straight line. It is made of a series of hollow cylindrical electrodes called drift tubes, inside an evacuated pipe. Alternate tubes are connected to opposite terminals of a high-frequency alternating p.d.
• Inside a drift tube there is no electric field, so the particle moves at constant velocity.
• In the gaps between tubes there is an electric field, and the particle is accelerated.
• While the particle is inside a tube, the p.d. reverses. So at every gap the tube behind repels the particle and the tube ahead attracts it, and it gains energy at every gap.The supply has a fixed frequency, so the particle must spend the same time (half a period) in each tube. Because its speed increases, each tube must be longer than the one before. For a tube where the particle's speed is v and the supply frequency is f:L = v / 2f
Near the speed of light the speed hardly increases, so the later tubes are almost the same length.Worked example: a proton crosses 20 gaps, each with a p.d. of 50 kV. It gains 50 keV per gap, so 20 × 50 keV = 1.0 MeV = 1.6 × 10-13 J.
If the proton moves at 2.0 × 106 m s-1 with f = 20 MHz, the tube length is L = 2.0 × 106 / (2 × 20 × 106) = 0.050 m.
• Inside a drift tube there is no electric field, so the particle moves at constant velocity.
• In the gaps between tubes there is an electric field, and the particle is accelerated.
• While the particle is inside a tube, the p.d. reverses. So at every gap the tube behind repels the particle and the tube ahead attracts it, and it gains energy at every gap.The supply has a fixed frequency, so the particle must spend the same time (half a period) in each tube. Because its speed increases, each tube must be longer than the one before. For a tube where the particle's speed is v and the supply frequency is f:
Near the speed of light the speed hardly increases, so the later tubes are almost the same length.Worked example: a proton crosses 20 gaps, each with a p.d. of 50 kV. It gains 50 keV per gap, so 20 × 50 keV = 1.0 MeV = 1.6 × 10-13 J.
If the proton moves at 2.0 × 106 m s-1 with f = 20 MHz, the tube length is L = 2.0 × 106 / (2 × 20 × 106) = 0.050 m.