Nuclear Density
As mentioned in previous article click here to read, the radius of the nucleus is very small but this small nucleus is very massive. Since nearly whole of the mass of atom is concentrated in this tiny nucleus, the nucleus has a very high density.
Mathematically
Consider a nucleus has ‘A’ nucleons in it.i.e. Number of protons + Number of neutrons is equal to A
Let mass of each nucleon \( m = 1.67 \times 10^{-27} \, \text{kg} \)
Mass of electron \( = 9.1 \times 10^{-31} \, \text{kg} \)
Mass of nucleus \( M = mA = 1.67 \times 10^{-27} A \, \text{kg} \)
If \( R \) is radius of nucleus,
\[ R = R_0 A^{1/3} \]
Taking \( R_0 = 1.2 \times 10^{-15} \, \text{m} \).
Then volume of the nucleus
\[ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 \]
\[ V = \frac{4}{3} \pi R_0^3 A \]
Density of the nucleus
\[ \text{Density} = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} \]
\[ \frac{M}{V} = \frac{1.67 \times 10^{-27} A}{\frac{4}{3} \pi R_0^3 A} \]
= (3 × 1.67 × 10-27) / (4 π R03) = (3 × 1.67 × 10-27) / (4 × 3.14 × (1.2 × 10-15)3)
= 2.29 × 1017 kg m-3 ... (1.9)
Thus density of nucleus is very high (≈ 1017 kg m-3) and it is independent of mass number 'A'.
conclusion
The experimental results show that the density of the nucleus is maximum at the centre and decreases to zero as we move radially outward.