Order-of-magnitude nuclear sizes: what is the approximate ratio of the volume of an atom to the volume of its nucleus?

Introduction:
Comparing atomic and nuclear dimensions illustrates why most of the atom is empty space. This scale difference underpins models of electron orbitals, scattering experiments, and the conceptual separation between nuclear and electronic phenomena.

Given Data / Assumptions:

• Typical atomic radius ~ 10^-10 m (or 10^-8 cm).
• Typical nuclear radius ~ 10^-15 m (or 10^-12 cm).
• Volumes scale with the cube of linear dimension.

Concept / Approach:
If r_atom / r_nucleus ≈ 10^5 in SI metres, the volume ratio is (10^5)^3 ≈ 10^15. However, using cgs common textbook radii (10^-8 cm vs 10^-12 cm) gives a linear ratio of 10^4 and a volume ratio of 10^12. Many introductory references and MCQs adopt this cgs-based comparison, yielding 10^12.

Step-by-Step Solution:
Choose representative radii in cm: atom ~10^-8 cm; nucleus ~10^-12 cm.Compute linear ratio: 10^-8 / 10^-12 = 10^4.Convert to volume ratio: (10^4)^3 = 10^12.

Verification / Alternative check:
Using SI radii (1 × 10^-10 m and 1 × 10^-15 m) gives 10^15; the discrepancy reflects the variability of quoted “typical” sizes. For standardized MCQs, 10^12 is a commonly accepted answer.

Why Other Options Are Wrong:

• 10^-12 and 10^-8: These represent tiny ratios, not the enormous atomic-to-nuclear volume ratio.
• 10^8: Too small compared to typical estimates.
• 10^5: This is close to the linear ratio (radius), not volume.

Common Pitfalls:
Mixing linear and volumetric ratios or confusing cgs and SI “typical” radii without cubing for volume.

Final Answer:
10^12