Size of the Nucleus

How the size is measured

• Rutherford postulated and established the nucleus; at his suggestion Geiger and Marsden scattered $\alpha$-particles from thin gold foils.
• Their result: for an $\alpha$-particle of kinetic energy $5.5$ MeV, the distance of closest approach to a gold nucleus is about $4.0\times10^{-14}$ m. Since the whole positive charge sits inside the nucleus, the nucleus must be SMALLER than $4.0\times10^{-14}$ m. This is an upper limit, not the radius itself.
• Rutherford's calculation assumed pure Coulomb repulsion between the $\alpha$-particle and the nucleus, and it reproduced the observed scattering. If $\alpha$-particles of higher energy are used they approach more closely, and at some point the short-range NUCLEAR force starts to act and the scattering departs from Rutherford's Coulomb prediction. The separation at which the deviation sets in is a measure of the nuclear size.
• More accurate sizes come from scattering FAST ELECTRONS (rather than $\alpha$-particles) off targets of various elements. Electron scattering probes the charge distribution of the nucleus, whereas $\alpha$-particles and similar heavy projectiles probe the nuclear matter distribution — which is why the two methods give slightly different radii.

The trajectory diagram, described in words (3-mark answer)

• Draw the target nucleus as a small positive charge at the centre. Draw a parallel beam of $\alpha$-particles incident from the left, at a range of perpendicular distances (impact parameters $b$) from the line through the nucleus.
• Each path is a HYPERBOLA curving away from the nucleus, because the Coulomb force between the two positive charges is repulsive.
• Large impact parameter: the particle passes almost undeviated, $\theta$ small.
• Smaller impact parameter: larger deflection; $\theta < 90^\circ$ for a moderate approach, $\theta > 90^\circ$ for a close approach.
• $b \to 0$ (head-on): the particle is turned straight back, $\theta \approx 180^\circ$.
• Label the perpendicular distance $b$ from the nucleus to the initial line of travel as the impact parameter, and the angle between the final and initial directions as the scattering angle $\theta$. Smaller $b$ gives larger $\theta$.

What the experiment tells us about size

• Only a very small fraction of the incident $\alpha$-particles — about 1 in 8000 — rebound through large angles. So very few make a near head-on approach.
• That implies the entire positive charge, and almost all the mass, is concentrated in an extremely small volume at the centre of the atom.
• In a head-on collision the $\alpha$-particle stops when all its kinetic energy has become electrostatic potential energy:

$$K = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d} \quad\Rightarrow\quad d = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}$$

• This distance of closest approach $d$ is an upper limit on the nuclear radius.
• The gold nucleus is roughly 50 times heavier than an $\alpha$-particle, so it stays essentially at rest during the encounter and the light $\alpha$-particle simply bounces off it — a single scattering event, not a succession of them.

The radius formula

• Electron-scattering experiments give, for a nucleus of mass number $A$,

$$R = R_0 A^{1/3}, \qquad R_0 = 1.2\times10^{-15}\ \text{m} = 1.2\ \text{fm}$$

• 1 fermi (fm) $= 10^{-15}$ m.
• Ratio form for comparisons:

$$\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3}$$

• Worked (board, 1 mark): the nuclear radius of ${}^{27}_{13}\text{Al}$ is $3.6$ fermi; find that of ${}^{64}_{29}\text{Cu}$.

$$R_{Cu} = R_{Al}\left(\dfrac{A_{Cu}}{A_{Al}}\right)^{1/3} = 3.6\left(\dfrac{64}{27}\right)^{1/3}$$

Since $\dfrac{64}{27} = \left(\dfrac43\right)^3$, the cube root is $\dfrac43$, so $R_{Cu} = 3.6\times\dfrac43 = 4.8$ fermi.

• Worked (NCERT 13.4): the ratio of the nuclear radii of ${}^{197}_{79}\text{Au}$ and ${}^{107}_{47}\text{Ag}$ is $\left(\dfrac{197}{107}\right)^{1/3} = (1.841)^{1/3} \approx 1.23$.

Nuclear density is independent of A — the standard 3-mark derivation

• Radius of a nucleus of mass number $A$: $R = R_0 A^{1/3}$, with $R_0$ a constant.
• Volume, treating the nucleus as a sphere:

$$V = \dfrac43\pi R^3 = \dfrac43\pi\left(R_0 A^{1/3}\right)^3 = \dfrac43\pi R_0^3 A$$

• Mass of the nucleus, with $m$ the average mass of one nucleon: $M \approx mA$.
• Hence the nuclear density

$$\rho = \dfrac{M}{V} = \dfrac{mA}{\dfrac43\pi R_0^3 A} = \dfrac{3m}{4\pi R_0^3}$$

• Both $m$ and $R_0$ are constants and $A$ has cancelled completely, so $\rho$ does not depend on the mass number: every nucleus has the SAME density. Numerically $\rho \approx 2.3\times10^{17}$ kg m$^{-3}$ with $R_0 = 1.2$ fm (CBSE's marking scheme sometimes quotes $2.97\times10^{17}$ kg m$^{-3}$, the same expression with $R_0 = 1.1$ fm).
• Immediate corollary for MCQs: the ratio of the nuclear densities of any two nuclei — mass numbers 64 and 125, or $X^{27}$ and $Y^{8}$, or any other pair — is $1:1$.

Why nuclear density is so enormous

• $\rho_{\text{nucleus}} \approx 2.3\times10^{17}$ kg m$^{-3}$ against $\rho_{\text{water}} = 10^{3}$ kg m$^{-3}$: a factor of about $10^{14}$.
• Reason to write: essentially the whole mass of an atom (more than $99.9\%$) is packed into a nuclear volume whose radius is $10^{4}$–$10^{5}$ times smaller than the atomic radius; the rest of the atom is empty space occupied only by very light electrons. Packing all the mass into so tiny a volume gives an enormous density.
• The nuclear radius is smaller than the atomic radius by a factor of about $10^{4}$, so the nuclear volume is about $10^{-12}$ of the atomic volume. If an atom were the size of a classroom, the nucleus would be the size of a pinhead.
• Because the density is the same for all nuclei, nuclei behave like drops of a liquid of constant density. Matter in a neutron star has a comparable density — such an object is effectively one giant nucleus.
• Note the contrast worth stating: nuclear density is independent of the size of the nucleus, but the mass density of an ATOM does depend on the atom, because the electron cloud's size varies.

Worked example (NCERT 13.1)

• Iron nucleus: mass $55.85$ u with $A = 56$, so mass $= 55.85\times1.66\times10^{-27} = 9.27\times10^{-26}$ kg.
• Volume $= \dfrac43\pi\left(1.2\times10^{-15}\times56^{1/3}\right)^3 = \dfrac43\pi(1.2\times10^{-15})^3\times56$.
• $\rho = \dfrac{9.27\times10^{-26}}{\dfrac43\pi(1.2\times10^{-15})^3\times56} = 2.29\times10^{17}$ kg m$^{-3}$.

Standard results

• $R = R_0 A^{1/3}$, $R_0 = 1.2$ fm; $R \propto A^{1/3}$; $V \propto A$; $\rho$ independent of $A$.
• $\rho \approx 2.3\times10^{17}$ kg m$^{-3}$, the same for all nuclei; ratio of nuclear densities always $1:1$.
• Distance of closest approach $d = \dfrac{2Ze^2}{4\pi\varepsilon_0 K}$; $\approx 4.0\times10^{-14}$ m for a $5.5$ MeV $\alpha$-particle on gold.
• Nuclear radius is about $10^{-4}$ of the atomic radius; nuclear volume about $10^{-12}$ of the atomic volume; nucleus holds over $99.9\%$ of the atomic mass.
• Electron scattering measures the charge radius; $\alpha$-scattering measures the matter radius, so the two differ slightly.