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1. The nuclear atom page 5
How big is the nucleus?
The nucleus is tiny – about 10–15 m across. Compare this with the size of an atom – about 10–10 m across – and you can see that most of the atom is empty space. The first estimates for the size of the nucleus came from the Rutherford experiment. However, not all nuclei are the same size and to get an accurate measurement, we use a version of the Rutherford experiment with electrons rather than alpha particles. First, let's see how Rutherford deduced the size of the nucleus.
Measuring the nuclear radius
In order to estimate the size of the nucleus, Rutherford used his analysis of the results to devise a model of what was happening. This model assumes that the nucleus is a point charge. However, the nucleus has some size (albeit tiny) and he could estimate this size by looking for readings that started to deviate from those predicted by his model. Let's look at this in more detail.
Deflecting alpha particles interactive graphic
Picture 1.9 Some alpha particles that pass close enough to a nucleus to be deflected.
The nucleus has a positive charge. This means that it will repel the positively charged alpha particles. Picture 1.9 shows the paths of alpha particle as they pass close to (or towards) the nucleus. The electrostatic repulsion makes them vere off course. Some of them are deflected straight back.

Rutherford assumed that the charge of the nucleus was concentrated at its centre and devised a formula for how many particles would be deflected in each direction. The results matched this formula extremely well – lending support to Rutherford's model.

He then reasoned that if an alpha particle were to penetrate the nucleus, then the alpha particle would be exposed to less charge. The effective charge acting on an alpha particle that penetrated the nucleus would be less than the total charge of the nucleus. This would mean that the alpha particle would be deflected less than expected. So, he hypothesised that if the alpha particles were penetrating the nucleus, the number being deflected by large angles would be less than predicted. However, the experiment revealed no such deviation, meaning that the alpha particles were not penetrating the nucleus. And therefore, he concluded that the nucleus must be smaller than the distance of closest approach of any of the alpha particles.

Interactive graphic showing KE to EPE
Picture 1.10 A graph of the potential energy of an alpha particle as it approaches a nucleus. As it gains EPE, it loses KE until it is stopped.
How close did they get?
The alpha particles that get closest to the nucleus are those that are heading straight for it. We can work out how close they get using their energy.The most energetic alpha particles that Geiger and Marsden used had an energy of 7.7 MeV. This is in the form of kinetic energy (KE). As they approach the nucleus, they lose KE and gain electrical potential energy (EPE). At the point at which they are stopped and turned round, they have no KE. So the electrical potential energy is equal to 7.7 MeV.
 

Electrical potential energy gained = KE lost

2 x 97 e2/4Pe0r = 7.7 MeV
 
r = 2 x 97 e/4Pe0 x 7.7 x 106
 
  = 8.98 x 109 x 2 x 97 x 1.6 x 10–19 /7.7 x 106
 
  = 3.6 x 10–14 m

He knew that the nucleus must be smaller than this. I.e. this was an upper limit for its radius.

Deviation from Rutherford formula
Picture 1.11 Rutherford formula and deviation
Getting closer to the nucleus
We can get closer to the nucleus using higher energy particles. The number scattered should be proportional to the inverse of the kinetic energy. Picture 1.11 is a graph showing this prediction for aluminium at an angle of 180° - i.e. the alpha particles are being scattered backwards. When the experiment is done, the results deviate from the prediction at an energy of 10.4 MeV. So, at this point, the electrons must be entering the nucleus.

Because we are considering back-scattered alpha particles, we can say that the alpha particles have lost all their KE and gained electrical potential energy when they are turned around. Therefore, the EPE of an alpha particle that just touches the nucleus is 10.4 MeV. This give a radius of 3.6 x 10–15 m.

Electron scattering
We get more accurate measurements with high energy electrons. These have the advantage that they themselves have no size (unlike alpha particles). Therefore, their own radius doesn't affect the results.

Low energy electrons are deflected in the same way as the alpha particles (even though the force is now attractive, the scattering angles are the same). The high energy electrons travel at about a fifth the speed of light, so there has to be a correction for their change of mass according to Special Relativity. Nevertheless, it is possible to derive a formula to predict the number of electrons scattered at a given angle and how this will change with the energy of the electrons.

Let's get measuring
In the 1950s, many physicists worked on measuring nuclear radii. It didn't take long to measure all the elements. They found that the radius of a nucleus follows a simple power law based on its atomic mass number.

             r = rA1/3          where r0 = 1.2 x 10–15 m

This is what you would expect if all nuclei were made of material with the same density and the amount of that material depended on the atomic mass number.

So, we can assume that all nuclei are made up of only protons and neutrons and the total number of these in the nucleus is equal to A, the atomic mass number.

Question 4
a) The volume of a sphere is 4/3pr3.
i. Show that the volume of a nucleus of atomic mass number A is 4Pr03 A.
ii. Assume that each of the nucleons has a mass equal to the mass of a proton (mp). Show that the density of the nucleus is 3mp/4Pr03.
iii. What does this tell you about the density of nuceli with different atomic mass numbers?

b) Calculate the density of nuclei. [mp = 1.67 x 10-27 kg]
c) What would be the radius of the Earth if it had this density? [Mass of the Earth is 5.972 x 1024 kg]
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