Question
Part A Find the speed of the proton. Express your answer to two significant figures and include the appropriate units.
Part B Find the speed of the alpha particle. Express your answer to two significant figures and include the appropriate units. Part C Find the proton’s distance of closest approach in a head-on collision with a gold nucleus. Express your answer to two significant figures and include the appropriate units. Part D Find the alpha particle’s distance of closest approach in a head-on collision with a gold nucleus. Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
Data: Energy of photon = Energy of alpha particle = 9 MeV = 9 * 1.6 x 10^-13 J = 1.44 x 10^-12 J (a) Kinetic energy, E = (1/2) m v^2 Speed of proton, v = ( 2 E / m ) = [ ( 2 * 1.44 x 10^-12 ) / 1.67 x 10^-27] = 4.2 x 10^7 m/s (b) Kinetic energy, E = (1/2) m v^2 Speed of alpha particle, v = ( 2 E / m ) = [ ( 2 * 1.44 x 10^-12 ) / ( 4 * 1.67 x 10^-27 ) ] = 2.1 x 10^7 m/s (c) Atomic no. of gold = 79 Charge of proton, q1 = 1.6 x 10^-19 C Charge of the gold nucleus, q2 = 79 * 1.6 x 10^-19 = 126.4 x 10^-19 C Kinetic energy = potential energy E = k q1 q2 / r r = k q1 q2 / E = 9 x 10^9 * 1.6 x 10^-19 * 126.4 x 10^-19 / 1.44 x 10^-12 = 1.3 x 10^-14 m (d) Atomic no. of gold = 79 Charge of alpha, q1 = 3.2 x 10^-19 C Charge of the gold nucleus, q2 = 79 * 1.6 x 10^-19 = 126.4 x 10^-19 C Kinetic energy = potential energy E = k q1 q2 / r r = k q1 q2 / E = 9 x 10^9 * 3.2 x 10^-19 * 126.4 x 10^-19 / 1.44 x 10^-12 = 2.6 x 10^-14 m Kinetic energy, E = (1/2) m v^2 Speed of alpha particle, v = ( 2 E / m ) = [ ( 2 * 1.44 x 10^-12 ) / ( 4 * 1.67 x 10^-27 ) ] = 2.1 x 10^7 m/s (c) Atomic no. of gold = 79 Charge of proton, q1 = 1.6 x 10^-19 C Charge of the gold nucleus, q2 = 79 * 1.6 x 10^-19 = 126.4 x 10^-19 C Kinetic energy = potential energy E = k q1 q2 / r r = k q1 q2 / E = 9 x 10^9 * 1.6 x 10^-19 * 126.4 x 10^-19 / 1.44 x 10^-12 = 1.3 x 10^-14 m (d) Atomic no. of gold = 79 Charge of alpha, q1 = 3.2 x 10^-19 C Charge of the gold nucleus, q2 = 79 * 1.6 x 10^-19 = 126.4 x 10^-19 C Kinetic energy = potential energy E = k q1 q2 / r r = k q1 q2 / E = 9 x 10^9 * 3.2 x 10^-19 * 126.4 x 10^-19 / 1.44 x 10^-12 = 2.6 x 10^-14 m Atomic no. of gold = 79 Charge of alpha, q1 = 3.2 x 10^-19 C Charge of the gold nucleus, q2 = 79 * 1.6 x 10^-19 = 126.4 x 10^-19 C Kinetic energy = potential energy E = k q1 q2 / r r = k q1 q2 / E = 9 x 10^9 * 3.2 x 10^-19 * 126.4 x 10^-19 / 1.44 x 10^-12 = 2.6 x 10^-14 m