# Derive an expression and numerical value for the distance of closest approach of an alpha to a...

## Question:

Derive an expression and numerical value for the distance of closest approach of an alpha particle to a gold nucleus, when the impact parameter is {eq}b=0 {/eq} (i.e., a head-on collision). Use conservation of energy, and take the alpha particle energy to be {eq}6 MeV (1 MeV = 1.602E-13 J) {/eq}.

a. How does this compare with the Bohr radius?

b. Why the difference, if any?

## Electric Potential Energy

The electric potential energy between two charged particles is given by:

{eq}U = \frac{Q_1 Q_2}{4 \pi \epsilon_0 r} {/eq}

Here {eq}Q_1 {/eq} and {eq}Q_2 {/eq} are the charges of the two particles, {eq}\epsilon_0 {/eq} is the electric constant ({eq}\approx 8.854 \times 10^{-12} {/eq} F/m, and r is the distance between them.

#### a)

Using conservation of energy we can say that at the point of closest approach all the alpha particles kinetic energy is converted to potential energy. Thus:

{eq}K = U {/eq}

Since we know {eq}K = 6 \: \text{MeV} = 9.61 \times 10^{-13} \: \text{J} {/eq}

we can solve for r as the distance of closest approach. An alpha particle has a charge of +2e, while a gold nucleus has a charge of +79e.

{eq}9.61 \times 10^{-13} = \frac{2 \times 79 \times e^2}{4 \pi \epsilon_0 r} {/eq}

{eq}r = \frac{2 \times 79 \times (1.609 \times 10^{-19})^2}{4 \pi (8.954 \times 10^{-12}) (9.61 \times 10^{-13})} = 3.83 \times 10^{-14} \: \text{m} {/eq}

Thus the distance of closest approach for the alpha particle is 3.83 {eq}\times 10^{-14} {/eq} meters.

#### b)

The Bohr radius for a hydrogen-like atom is:

{eq}a_0 = \frac{5.29 \times 10^{-11} \: \text{m}}{Z} {/eq}

Here Z is the number of protons in the nucleus. For a gold nucleus this gives {eq}6.70 \times 10^{-13} {/eq} meters. This is about 20 times larger than the distance of closest approach of our alpha particle.

#### c)

The Bohr radius of a hydrogen-like atom describes the most likely distance an electron would be from the nucleus, and is derived from quantum-mechanical models of the interaction. Seeing as how our derivation of the closest approach of our alpha particle was using purely classical physics to describe the potentials and kinetic energies it is not surprising that the distance we found was not very similar to the Bohr radius. Furthermore the Bohr radius describes the most likely position of an electron orbiting around the nucleus, seeing as how we are modelling an alpha particle which is not orbiting the nucleus but instead is colliding with it, it is fairly reasonable that the Bohr radius is not applicable to this case since it models an entirely different scenario. 