Question

(a) calculate the speed of a proton after it accelerates from rest through a potential difference of 215 v. m/s (b) calculate the speed of an electron after it accelerates from rest through a potential difference of 215 v. m/s

Answer 1

The proton differs from the electron in sign although they have the same value. Like the electron, a proton will gain 215 electron-volts of eV in Kinetic energy. So 1.602Ă—10^-19 J * 215 = 344.43 * 10^(-19) J. But K. E. = mv^2 / 2, so v^2 = 2 * K.E/m. The mass of a proton is 1.673 * 10^-27 kg. So v = âš(2 * 344.43 * 10^(-19))/1.673Ă—10^-27 = 688.86 * 10^(-19)/1.673Ă—10^(-27) = 411.75 * 10^(-19-(-27)) = âš411.75 * 10^(8) = 202196.56 Also for the electron we have v^2 = 2 * K.E/m but here mass, m, = 9.109 * 10^-31 kg. So we have v = âš(2 * 344.43 * 10^(-19)) / 9.109 * 10^-31 = 688.86 * 10^(-19)/ 9.109 * 10^-31 = 75.624 * 10^(-19 - (-31)) = 75.624 * 10^(21) and v = 2.749 * 10^11