Question

Cyclotrons are sometimes used to make technetium-99m, the most widely employed isotope in nuclear medicine. This is done by bombarding a molybdenum target with protons to drive the nuclear reaction 100Mo(p,2n) 99mTc. The reaction notation means one bombards molybdenum having a mass number of 100 with protons to yield two neutrons and technetium-99m. The reaction rate is maximized at a proton energy of 17 MeV. You have a magnet that will produce a uniform field of up to 1.5 T over a circular area 1.2 m in diameter. You design the dees such that the proton beam leaves the magnetic field at a radius of 0.6 m to strike the 100Mo target.

(a) What value of magnetic field should you use to maximize the reaction rate?

(b) At what frequency must the cyclotron operate?

(c) What is the equation describing the radius of the proton path as a function of the number of accelerations, N, (that is the number of times the proton passes between the dees), the potential between the dees, V, the magnitude of the magnetic field, B, and the charge-to-mass ratio of the proton q/m

(e) If the potential between the dees is 90 kV what is the approximate length of the path traveled by the protons during the acceleration process?

Answer 1

Part a: The radius of the circular path is: mv OR >Bmp C q B QR The total energy is: K= -mv2 The speed: v= Then: m 2K_1 J2mK B = — qR QR Using the given data: K = 17 MeV = 17 X 106(1.6 X 10-19) = 2.72 * 10-12 ] Then: B =- V 1.67 * 10-27 X 2.72 10-12 1.6 X 10-19 x 0.6 -- B= 0.702 T Part b: The angular frequency is: _v_qB (1.6 X 10-19) (0.702) rad R m 1.67 * 10-27 = 6.73 * 107 tad Part c: The radius is given by: | = V qВ After the 1 acceleration: 2qV v m After N acceleration: m R= V = m qBm Parte: The length is: Using the given data: 20 103 L = 2 * 1.67 * 10-27 * 90 X U-19 = 0.388 m 0.702