Question

An alpha particle has a mass of 6.65 x 1027 kg and a charge of +3.2 x 1019 C (it is the same as a Helium nucleus). A large, uncharged parallel plate capacitor of capacitance 0.01 nF sits near a radioactive sample that emits alpha particles. A very slow-moving alpha particle ends up inside the capacitor exactly in between the two plates (equidistant from each plate). A 1.5V battery is then connected across the capacitor so that the positive terminal is connected to one plate and the negative terminal is connected to the opposite plate. Describe the motion of the alpha particle. Estimate its change in kinetic energy and its maximum velocity before it collides with one of the plates.

Answer 1

"Given data V = 1.5 Charge q = 3.2 times 10^-19 c Delta v = 0.75 v Therefore Delta v = v/2 = 1.5/2 Delta v = 0.75 volts We know Delta KE = q Delta v -eq^n (1) Substitute all values in eq^n (1) Delta KE = (3.2 times 10^-19 c) (0.75 v) Delta KE = 2.4 times 10^-19 Delta KE = 2.4 times 10^-19 J (or) 1.5 ev - eq^n (2) Delta KE = mv^2/2 Substitute Delta KE from eq^n (2) 24 times 10^-19 J = (6.65 times 10^-27)/2 v^2 \r\n 4.8 times 10^-19 J = (6.65 times 10^-27) v^2 Here we have to find ""v"" v^2 = 4.8 times 10^-19/6.65 times 10^-27 v^2 = 7.2180 times 10^-47 v = squareroot 7.218 times 10^-47 v = 8.49 times 10^3 m/s (or) we know 1/2 m v^2 = q (A v) v = squareroot q (Delta v)/m = squareroot (3.2) times 10^-19. times 1.5 v"