1. History

Question 1

Plum-Pudding Model

Answer 1

• prior to the discovery of the nucleus, the most widely accepted model of the atom was the plum-pudding model
• electrons embedded in a diffuse cloud of positive charge
• a reasonably successful model though it failed to explain emission spectra

Question 2

Rutherford Experiment

Description

Answer 2

• alpha particles (helium nuceli) fired at a thin piece of gold foil
• deflections were observed using a fluorescent screen

Question 3

Rutherford Experiment

Conservation of Energy

Answer 3

-consider the scattering process of a moving alpha particle off a stationary target:
1/2 mα vo² = 1/2mα vα² + 1/2mt vt²
-rearranging:
vo² = vα² + mt/mα * vt²

Question 4

Rutherford Experiment

Conservation of Momentum

Answer 4

-consider the scattering process of a moving alpha particle off a stationary target:
mαvo = mαvα + mt*vt

Question 5

Rutherford Experiment

Combining Conservations of Energy and Momentum

Answer 5

vt² (1 - mt/mα) = 2vα . vt =

2 |vα| |vt| cosθ

Question 6

Rutherford Experiment

target mass less than α mass

Answer 6

-if the target mass is much less than the α mass (e.g. α scattered off an electron), mt/mα«1 then:
vt² ≈ 2 |vα| |vt| cosθ
-this makes cosθ > 0 (θ

Question 7

Rutherford Experiment

target mass greater than α mass

Answer 7

-if the target mass is much greater than the α particle mass, the α particle can recoil sharply backwards in the direction it originally came from

Question 8

Rutherford Experiment

Observations and Implications

Answer 8

• Rutherford found that the α particles occasionally recoiled from the gold foil
• the plum pudding model was unable to account for this
• instead we need an atomic model with high-mass objects
• since the large angle deflections were rare, the heavy object must be small

Question 9

Nuclear Model of the Atom

Answer 9

• heavy dense nucleus
• electrons in large orbitals
• mostly empty space

Question 10

Impact Parameter

Answer 10

• consider an α particles trajectory as it collides with a gold nucleus
• the perpendicular displacement of the initial trajectory from perfect head-on collision is called the impact parameter, b

Question 11

Large b

Answer 11

-for larger b, the Coulomb repulsion is low

Question 12

Small b

Answer 12

-for very small b, the α particle moves through the nucleus and is therefore subject to a smaller effective charge

Question 13

b of order of the nuclear radius

Answer 13

-for b of the order of the gold nucleus’s radius, the deflection is maximum

Question 14

Calculating Deflection

Approximation

Answer 14

-in an approximate calculation we can take the Coulomb repulsion to act in a direction perpendicular to the initial trajectory, and only over a distance b
-so for an α particle velocity v, we assume the force only acts for a time:
Δt = b/v

Question 15

Calculating Deflection

Coulomb Force

Answer 15

F = 1/4πεo * (2e)(Ze)/b²

• where 2e is the charge on the α particle
• and Ze is the charge on the gold nucleus

Question 16

Calculating Deflection

Change in Momentum

Answer 16

-the deflection angle θ in radians is approximately
θ = Δp/p , but F = Δp/Δt
so:
θ = 1/4πεo * 2Ze²/b² * b/v * 1/mv
-rearrange for b:
b =  1/4πεo * 2Ze²/mv² * 1/θ

Question 17

Calculating Deflection

Nuclear Radius

Answer 17

-from the data we can see that maximum deflection is of order 1 radian
-subbing in the other numbers we get:
B = R ≈ 10^(-14) m
-note that this is much smaller than the atomic radius

Question 18

Rutherford Scattering Formula

Answer 18

N(θ) = (1/4πεo)² * NiρLZ²e^4/4r²E²sin^4(θ/2)
-for point-particle targets where:
N(θ) = no.of particles scattered at θ
Ni = number of incident particles 
ρ = density of target
L = target thickness
r = target-to-deflector distance
E = kinetic energy of α particle

Question 19

When does the Rutherford scattering formula fail?

Answer 19

• at sufficiently high energy levels the α particles will penetrate the nucleus
• at this point, the nucleus no longer behaves as a point-particle and the Rutherford formula fails

Question 20

Rutherford Scattering Formula and Protons

Answer 20

• a similar breakdown of the Rutherford formula occurs with protons as with nuclei
• this demonstrates that protons are not fundamental

Question 21

Strong Nuclear Force

Answer 21

• since Rnucleus ≈ 10^(-15)-10^(-14) m and all the atom’s positive charge is contained in such a small region, there must be a huge attractive force to overcome the Coulomb repulsion
• this is the strong nuclear force
• the strong force is only a short-range force and its effects are not seen at scales larger than the nuclear radius

Question 22

The Strong Force and the Rutherford Scattering Experiment

Answer 22

• the strong force is also present in α particles and this can confuse matters in the Rutherford scattering experiment
• for this reason the most accurate versions of the experiment use electrons instead as they are not subject to the strong nuclear force