Momentum, Kinetic Energy and the Discovery of the Neutron

 

 


 

before elastic collision

 

 

Consider two masses, m1 and m2, about to collide head on. If their collision is perfectly elastic, the total momentum and kinetic energy of the system will be conserved. (in an inelastic collision--the masses lock together--- only momentum is conserved)

 


after elastic collision

 

 

 

 

Since kinetic energy is conserved:

0.5 m1v12 + 0.5 m2v22 = 0.5 m1v1' 2 + 0.5 m2v2' 2  or

0.5 m1v12 - 0.5 m1v1' 2 = 0.5 m2v2' 2 - 0.5 m2v22

First we will divide through by 0.5, then take out a common factor of m1 and then we will factor the difference of squares:

m1(v1 - v1')(v1 + v1') = m2(v2 - v2')(v2 + v2') or

(m1v1 -  m1v1')(v1 + v1') =   ( m2v2 -  m2v2') (v2 + v2')             

 But the terms in bold are the changes in momentum experienced by each ball. Since whatever momentum is gained by one ball is gained by the other, these terms cancel and we are left with v1 + v1' = v2 + v2'.                                   (1)

                             

Momentum:               

m1v1 + m2v2 = m1v1'  + m2v2'                                                          (2)

But since, from rearranging equation(1), v2' = v1 + v1' - v2 , it can be substituted this into equation(2):

 m1v1 + m2v2 = m1v1'  + m2(v1 + v1' - v2 )

Solving for v1' we obtain

                                                    (3)

Similarly if we isolate v2' in equation (1) and substitute it into eqn (2) and then solve for v2' we obtain:

                                                    (4)

What is all this have to do with the discovery of the neutron? When alpha particles (equivalent to a helium nucleus)  from the radioactive element polonium bombarded beryllium, Be, invisible particles of mass m1 were emitted. These struck either N nuclei or protons initially at rest, and Chadwick then measured the resulting velocities of the moving nuclei, v2'

Since the nuclei were initially at rest, v2 = 0. So equations (3) and (4) simplify to:

         (5)   and                                    (6)

Chadwick compared the speed v2' for nitrogen to that of hydrogen when each collided with the mysterious component(neutron) emitted from beryllium’s nucleus: Note that N is 14 times more massive than H(corresponding to subscript 2), and the subscript 1 refers to the neutron.

and

This ratio, according to his experiments, turned out to be 7.5. From this he concluded that m1, the mass of the unknown particle emitted by beryllium, was about the same as that of the proton, m2. Chadwick had found an essential property of the previously unknown neutron by applying the conservation of momentum and kinetic energy principles to subatomic particles. Note that in the case of a collision with a stationary proton, the neutron should end up with a velocity of zero (see equation(5) ). This is similar to hitting a cue ball head on with another ball. The cue ball will stay put after the collision. Here is Chadwick's original paper.

References

Serway, Richard A. Physics for Scientists and Engineers. Saunders. 1983.

Haber-Schaim, Uri and al. PSSC Physics 3rd Edition. Heath. 1971.

Martindale,  David G. and al. Physics: A Senior Course. Heath Canada. 1986.

Giancoli, Douglas C. Physics. Prentice Hall. 1995.