Nuclear Volume and Nearly Constant Density
Using M approximately A m_n and R = R0 A^(1/3), a spherical nucleus has volume (4/3)pi R0^3 A and density 3m_n/(4pi R0^3), independent of A within the liquid-drop approximation.
Why this shows up in the exam
Comparing densities of two nuclei · Estimating the order of nuclear density · Relating nuclear mass to volume
Learn the idea
Because nuclear mass and volume both scale with A, nuclear density is nearly independent of mass number. A larger nucleus contains more nucleons but also occupies proportionally more volume. Their ratio stays almost fixed, like adding equally packed blocks without changing the packing density.
🧠 Memory hook: Mass follows A and volume follows A, so density cancels A.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- V = (4/3) pi R0³ A — spherical nuclear volume after using the empirical radius law
- rho approximately 3 m_n / (4 pi R0³) — nearly constant nuclear density when proton and neutron masses are approximated by one nucleon mass
How to approach it
- 1Substitute R cubed before simplifying
- 2Cancel A symbolically
- 3Use about 10¹⁷ kg m⁻³ as an order check
Common slip-ups that cost marks
- •Claiming heavier nuclei must have greater nuclear density
- •Using atomic size instead of nuclear radius
- •Forgetting the spherical-volume factor
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.
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