- Z1 is less than Z2 in the diagram above.
- A different charge causes different forces (F = kQq/r2), given b (r) is kept constant. Therefore a different charge causes a different scatter angle.
- It can be shown that Nu a Z2 by using:
- Newton's laws
- Coulomb's laws
- the geometry of the scattering interaction.
- Once these calculations are made, they can be verified experimentally by comparison with the scattering data collected by firing alpha particles at different foils made of different atoms (thickness of the foil kept constant).
- This means that if we didn't know the charge on the nucleus, we can calculate it from the scattering data.
THE SIZE OF THE NUCLEUS
- Back scattered particles allow us to get an upper limit to the size of the nucleus.
- Alpha particles of mass m and charge q, accelerate through a potential difference of V. For this circumstance we can write
qV = Eko = (1/2)m(vo)2 = Etot
- As the particle gets closer to the nucleus, the kinetic energy changes to potential energy or:
Etot = Ek + Ep = Eko = a constant
- A direct approach will stop the alpha particle at a distance ro (the distance of closest approach). Therefore at closest approach:
Etot = Ep = Eko = (1/2)m(vo)2
- Potential energy is calculated by calculating work. Two unlike charges attract with a force given by Coulomb's law.
F = (kQq)/(r2)
- To separate them, a Coulomb's force is applied over the distance between them. That is:
W = Fd (d = r)
W = (kQq/r2)r
Ep= (kQq)/r
- Given the equation above for electric potential energy, and the fact that the kinetic energy gets transformed to potential energy, we can write:
kQq/ro = (1/2)m(vo)2
or ro = (2kQq)/(m(vo)2)
- Given that an alpha particle is accelerated through a potential difference of 2.64 X 106 V (the mass = 6.6 X10-27 kg and q = 3.2 X 10-196 C), then:
qV = (1/2)m(vo)2
(3.2 X 10-19 C)(2.64 X 106 V) = (1/2)(6.6 X 10-27kg)(vo)2
vo = 1.6 X 107 m/s
- A gold nucleus has a charge of 1.3 X 10-17 C. The distance of closest approach is therefore:
ro = (2(8.99 X 109Nm2/C2)(3.2 X 10-19C)(1.3 X 10-17C))/((6.6 X 10-27kg)(1.6 X 107m/s)2)
or ro = 4.2 X 10-14m
- This gives an upper limit for the size of a gold atom. If an alpha particle comes closer than
- 4.2 X 10-14 m, it may contact the nucleus, and scattering would then not conform to the predictions based on Coulomb's law.
- Measurements (X-ray diffraction) gives us the distance between 2 gold atoms to be
- 2.5 X 10-10 m. This means that the radius of a gold atom is 1.2 X 10-10 m.
- The ratio of the volume of the atom to the volume of the nucleus, shows that the atom is mainly empty.
January 20, 2014