[<< wikibooks] AQA A-Level Physics/Nuclear Radius
When Ernest Rutherford first proposed the nuclear model of the atom in 1911,it enabled us to deduce the possible size for the atom. In Rutherford's experiment he fired a narrow beam of alpha particles (all with equal quantities of kinetic energy) in a vacuum to probe the structure of the atom. A thin strip of metal foil was placed in the path of the beam. Whilst most of the particles passed straight through the foil with very little or no deflection, approximately 1 in 10000 alpha particles were deflected by more than 90 degrees. The foil itself had to be very thin otherwise the particles could have been deflected more than once.
For this single scattering by a foil strip that has 'n' layers of atoms, the probability that an alpha particle is deflected by a given atom is therefore around 1 in 10000n. This probability does however depend on the effective area of cross-section of the nucleus to that of the atom. As a result, we can deduce the following...
For a nucleus of diameter, 'd', in an atom of diameter, 'D', the area ratio is equal to;

1
4

π

d

2

1
4

π

D

2

{\displaystyle {\frac {{\frac {1}{4}}\pi d^{2}}{{\frac {1}{4}}\pi D^{2}}}}

Therefore...

d

2

=

D

2

10000
n

{\displaystyle d^{2}={\frac {D^{2}}{10000n}}}

A standard value of

n
=

10

4

{\displaystyle n=10^{4}}
gives

d
=

D
10000

{\displaystyle d={\frac {D}{10000}}}

Electron scattering
This is a good way to estimate the size of a nucleus but a more accurate way of measuring the nucleus is electron scattering. In electron scattering, electrons are fired in a beam at a sheet of the target material. The de Broglie wavelength of an electron is about 10−15m. This means they are diffracted by the target material with the following relationship: 0.61λ/R=sinθ_min