Size of the Nucleus: Derivation of Radius and Nuclear Density

SIZE OF THE NUCLEUS

Get an expression for the radius of a nucleus in terms of mass number (A).

The first experimental determination of the size of a nucleus was made from the results of Rutherford scattering of α-particles. Distance of closest approach was found to be $4 \times 10^{-14}$ m for 5.5 MeV energetic α-particles. This fact indicated that the size of nucleus should be less than $4 \times 10^{-14}$ m. For α-particles having kinetic energy more than 5.5 MeV, the distance of closest approach will be smaller. For kinetic energy more than 5.5 MeV, attractive nuclear forces start affecting the Coulomb's repulsive force between α-particles and gold nucleus. Size of nucleus can be measured by using fast electrons instead of α-particles for the scattering experiment.

The size of nucleus was found to vary linearly with mass number (A). Since nucleus is supposed to be spherical, having radius ,

$V=\left(\frac{4}{3}\pi R^3\right)\propto A$

Therefore,

$R \propto A^{1/3}$

or

$R = R_0 A^{1/3}$

$R_0$ is an empirical constant for all nuclei and its value is calculated as

$R_0 = 1.2 \times 10^{-15}\ \text{m}$

As $10^{-15}m = 1 fermi (fm)$

So,

$R_0 = 1.2\ fm$

Nuclear Density

The mass per unit volume of a nucleus is called nuclear density.

$\text{Nuclear density},\ \rho = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}}$

$\rho= \frac{\text{Mass of Proton or Neutron × mass number}}{\text{Volume of nucleus}}$

$\rho = \frac{(1.66 \times 10^{-27})A}{V}......(i)$

Consider a nucleus with mass number A and radius R .

Volume of nucleus,

$V=\left(\frac{4}{3}\pi R^3\right)$

Using,

$R=R_0A^{1/3}$

where

$R_0=1.2\times10^{-15}\ \text{m}$

we get,

$V=\left(\frac{4}{3}\pi R_0^3A\right)$

$=\frac{4}{3}\pi(1.2\times10^{-15})^3A......(ii)$

From equation (i) and (ii), we get,

$\rho=\frac{(1.66\times10^{-27})A}{\frac{4}{3}\pi(1.2\times10^{-15})^3A}$

$=\frac{3\times1.66\times10^{-27}}{4\times3.14\times(1.2\times10^{-15})^3}$

$\rho = 2.38\times10^{17}\ \text{kg m}^{-3}$

Conclusion

Nuclear density is a constant and is independent of the mass number of a nucleus.