Question

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 9.7 MeV (= 9.7 × 106 eV) far from the gold nucleus. Take the mass of a nucleon (a proton or a neutron) to be 1.7 × 10-27 kg and assume that the mass of a nucleus equals the sum of the masses of its constituent nucleons. Assume also that all speeds are low compared to the speed of light.

50% Part (a) Find the final momentum of the alpha particle (with its sign), long after it interacts with the gold nucleus, in units of kg⋅m/s.

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.

Answer 1

let m1 = 4*m
KE1 = (1/2)*m1*u1^2

u1 = sqrt(2*KE1/m1)

= sqrt(2*9.7*10^6*1.6*10^-19/(4*1.7*10-27))

= 2.136*10^7 m/s

m2 = 197*m
u2 = 0

a) speed of alfa particle after the collision,

v1 = (m1 - m2)*u1/(m1 + m2)

= (4*m - 197*m)*2.136*10^7/(4*m + 197*m)

= -2.05*10^7 m/s

momentum of alfa particle, P1f = m1*v1

= 4*m*v1

= 4*1.7*10^-27*(-2.05*10^7)

= -1.4*10^-19 kg.m/s   <<<<<<<<<<<<---------------------Answer

b) speed of gold nucleus after the collision,

v2 = 2*m1*u1/(m1 + m2)

= 2*4*m*2.136*10^7/(4*m + 197*m)

= 8.50*10^5 m/s

momentum of gold nucleus, P2f = m2*v2

= 197*m*v2

= 197*1.7*10^-27*8.5*10^5

= 2.8*10^-19 kg.m/s <<<<<<<<<<<<---------------------Answer