Class Management Help HW 19 Begin Date: 8/9/2017 12:00:00 AM-Due Date: 1172017 1
ID: 1776098 • Letter: C
Question
Class Management Help HW 19 Begin Date: 8/9/2017 12:00:00 AM-Due Date: 1172017 11:59:00 PM End Date: 1/10 2017 11:59:00 PM (14%) Problem 6: In order to investigate the structure of atoms, Ernest Rutherford performed his famous experiment, in which he bombarded gold atoms with alpha particles and studied the scattering of the alpha particles. Imagine that an alpha particle (a helium nucleus, consisting of two protons and two neutrons) is initially moving along the x-axis in the positive direction straight toward an initially stationary gold nucleus (containing 79 protons and 118 neutrons) and all subsequent motion takes place along the x-axis. The alpha particle starts with kinetic energy of 8.9 MeV ( 8.9 × 100 eV) far from the gold nucleus. Take the mass of a nucleon (a proton or a neutron) to be 1.7 x 10-27kg and assume that the mass of a nucleus cquals the sum of the masses of its constituent nucleons. Assume also that all speeds are low compared to the speed of light. 50% Parl (a) Find Ihe linal momentum ofthe alpha parucle (with lls sign) long aner il neracls with the gold nucleus in unis of kgms. Grade Summary Deductions Potential Palpha 0% 100% sin) cotanasin acos) atanacotan)sinh cosh cotanh cosO Submissions Attempts remaining: 4 (0% per attempt) detalled view 0 egreesRadians Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 3 Feedback: 0%-deduction per feedback. 50% Part (b) Find the final momentum of the gold nucleus (with its sign), long after in interacts with the alpha particle, in units of kg·m/s.Explanation / Answer
Given that kinetic energy of the alpha particle is KE = 8.9 MEv = 8.9*10^6*1.6*10^-19 = 1.424*10^-12 J
mass of the alpha particle is m = 4*1.7*10^-27 = 6.8*10^-27 Kg
initial momentum is pi = sqrt(2*k*m) = sqrt(2*1.424*10^-12*6.8*10^-27) = 1.39*10^-27 kg m/sec
initial velocity is vi = pi/m = 1.39*10^-27/(6.8*10^-27) = 0.204 m/sec
suppose if the collision is elastic
then
final speed of alpha particle is V1f = (m1-m2)*V1i/(m1+m2)
m2 is the mass of the gold nucleus = (79+118)*(1.7*10^-27) = 334.9*10^-27 kg
then
V1f = (6.8-334.9)*0.204/(6.8+334.9) = -0.195 m/sec
final momentum is palpha = 6.8*10^-27*(-0.195) = - 1.326*10^-27 kg m/sec
final momentum of the gold nucleus
Pgold = 2*m1*v1i/(m1+m2) = 2*6.8*10^-27*0.204/((6.8+334.9)*10^-27) = 0.0082 m/s = 8.2*10^-3 m/s