Q.Obtain approximately the ratio of the nuclear radii of the gold isotope 79197Au and the silver isotope 47107Ag.
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Nuclear Density: Why All Nuclei Are Almost Equally Dense
Imagine you have a bag of marbles. If you pack them tightly, the density of the bag depends only on the marbles themselves — not on how many you put in. The nucleus behaves the same way. That's the core idea.
The Intuition
An atom's nucleus is made of protons and neutrons (collectively called nucleons). These nucleons are held together by the strong nuclear force, which is extremely short-ranged. Think of it like magnets: each nucleon only "feels" its immediate neighbours. So adding more nucleons doesn't compress the inner ones — it just adds a new layer on the outside.
This means the nucleus grows in volume proportionally to the number of nucleons. Double the number of nucleons, double the volume. And since mass also doubles, the density stays constant.
The Precise Statement
The nuclear radius R is experimentally found to follow:
R=R0A1/3
where:
- A = mass number (total protons + neutrons)
- R0≈1.2×10−15 m (a constant, about 1.2 femtometres)
R=R0A1/3
This is the nuclear radius formula. It's not a guess — it comes from scattering experiments where high-energy electrons or alpha particles bounce off nuclei.
Deriving the Density
The nucleus is roughly spherical, so its volume is:
V=34πR3=34π(R0A1/3)3=34πR03A
Notice: volume is directly proportional to A. The mass of the nucleus is approximately m≈A×(1.67×10−27 kg) (mass of one nucleon). So density:
ρ=volumemass=34πR03AA×mnucleon=34πR03mnucleon
The A cancels out completely. The density is a constant — independent of the nucleus size.
Nuclear density is independent of mass number A. All nuclei have approximately the same density.
The Numerical Value
Plug in the numbers:
- mnucleon≈1.67×10−27 kg
- R0≈1.2×10−15 m
ρ=34π(1.2×10−15)31.67×10−27≈2.3×1017 kg/m3
That's about 230 million tonnes per cubic centimetre. To put it in perspective: a sugar-cube-sized piece of nuclear matter would weigh as much as 230 million cars.
| Object | Density (kg/m³) |
|--------|-----------------|
| Water | 103 |
| Earth (average) | 5.5×103 |
| White dwarf star | 109 |
| Atomic nucleus | 2.3×1017 |
Why This Matters
This constancy of density tells us something profound: the strong nuclear force saturates. Each nucleon only interacts with its nearest neighbours, not with the whole nucleus. If the force were long-range (like gravity), density would increase with size. It doesn't — so the force is short-range. …
Why this formula?
Why Nuclear Density is Constant — The Reasoning
The most striking result about nuclear density is that it is roughly the same for all nuclei, regardless of size. This is not obvious — why wouldn't a larger nucleus be denser? The answer lies in how nuclear force works and how nucleons pack together.
Step 1: The nuclear volume formula
Experiments show that the radius of a nucleus is given by:
R=R0A1/3
where R0≈1.2×10−15 m (1.2 fm) and A is the mass number (total number of protons + neutrons).
The A1/3 dependence is the key. It means volume grows linearly with A, not faster.
Why A1/3? Because nucleons are packed as tightly as possible — like spheres in a close-packed arrangement. If you double the number of nucleons, you need to double the volume, so the radius must increase by 21/3.
Step 2: Volume from the radius
Assuming the nucleus is a sphere:
V=34πR3=34π(R0A1/3)3=34πR03A
Notice: the A1/3 cube gives A directly. So volume is proportional to A.
Step 3: Mass of the nucleus
The mass of the nucleus is approximately:
M≈A⋅mnucleon
where mnucleon≈1.67×10−27 kg (the average mass of a proton or neutron). The small mass defect from binding energy is negligible for this calculation.
Step 4: Density
Nuclear density ρ is mass divided by volume:
ρ=VM=34πR03AA⋅mnucleon=34πR03mnucleon
The A cancels completely. Nuclear density is independent of the nucleus size.
Step 5: The numerical value
Plugging in the numbers:
ρ=34π(1.2×10−15)31.67×10−27≈2.3×1017 kg/m3
ρnuclear≈2.3×1017 kg/m3 …
The key idea is that nuclear radius follows the empirical formula R=R0A1/3, where A is the mass number and R0 is a constant (~1.2 fm). This means the ratio of radii depends only on the cube root of the mass numbers.
Step 1: Write the radius for each nucleus:
RAu=R0(197)1/3, RAg=R0(107)1/3.
Step 2: Take the ratio: …
Nuclear radius scales as R∝A1/3, so the ratio of radii for gold (A=197) and silver (A=107) is 3197/107≈1.23.
The key idea is that nuclear density is roughly constant for all nuclei — a fact known from scattering experiments. Since mass number A is proportional to volume, and volume scales as R3, we get R∝A1/3. This is one of the most reliable scaling laws in nuclear physics.
Let’s work through it.
- The nuclear radius formula For any nucleus, the radius is given by
R=R0A1/3
where R0≈1.2×10−15m (a constant). The exact value of R0 cancels out when we take a ratio, so we don’t need it here.
- Write the ratio For gold: RAu=R0(197)1/3 For silver: RAg=R0(107)1/3 Dividing:
RAgRAu=(107)1/3(197)1/3=(107197)1/3
- Approximate the fraction 197/107≈1.841. Now take the cube root. A quick mental check: 1.23=1.728, 1.253=1.953. So the cube root lies between 1.2 and 1.25. More precisely, 1.233=1.23×1.23×1.23=1.5129×1.23≈1.861 — very close to 1.841. So (107197)1/3≈1.23 …
Method: Using the Radius–Mass-Number Relation
The key idea is that nuclear radius depends only on the mass number A, not on the atomic number Z. The standard empirical formula is:
R=R0A1/3
where R0≈1.2×10−15 m is a constant (the same for all nuclei). This means the ratio of two nuclear radii depends only on the ratio of their mass numbers.
Steps
-
Write the radius for each nucleus
For gold: RAu=R0(197)1/3
For silver: RAg=R0(107)1/3
-
Take the ratio — the R0 cancels immediately:
RAgRAu=(107)1/3(197)1/3=(107197)1/3
-
Simplify the fraction
107197≈1.841
-
Take the cube root …
Common Mistakes in Nuclear Radius Ratio Problems
Students typically make the same few errors when calculating the ratio of nuclear radii from mass numbers. Here is what goes wrong and how to fix each.
Mistake 1: Forgetting the Cube Root
The most frequent error is treating the radius as directly proportional to the mass number A, rather than to A1/3.
The nuclear radius formula is:
R=R0A1/3
where R0 is a constant (about 1.2×10−15 m). So the ratio of radii is:
R2R1=(A2A1)1/3
Students often write R2R1=A2A1 instead, which gives a wildly wrong answer.
Never write R∝A. The correct proportionality is R∝A1/3. The cube root is non-negotiable.
How to avoid: Every time you see "nuclear radius" and "mass number" together, immediately write R=R0A1/3 before doing anything else. Then take the ratio — the R0 cancels, and you are left with the cube root of the mass number ratio.
Mistake 2: Confusing Mass Number with Atomic Number
The formula uses A (mass number = protons + neutrons), not Z (atomic number = protons only). For gold, A=197; for silver, A=107. The atomic numbers (79 and 47) are irrelevant here.
The nuclear radius depends on the total number of nucleons, not just protons. Ignore Z in this calculation.
How to avoid: Circle the mass numbers in the isotope notation. The superscript (top-left) is A; the subscript (bottom-left) is Z. Use only the superscript.
Mistake 3: Inverting the Ratio
Students sometimes write RAgRAu=(197107)1/3 instead of (107197)1/3. Gold has more nucleons, so its nucleus is larger — the ratio must be greater than 1.
How to avoid: Check your common sense. Gold (A=197) is heavier than silver (A=107), so its radius should be larger. If your ratio comes out less than 1, you have inverted it.
Mistake 4: Not Simplifying the Cube Root
Students often leave the answer as (107197)1/3 without approximating. The question asks for an approximate ratio — you need a numerical value. …
- COMEDK 2026Set 2026-A1 markMCQQ.If the ratio of the nuclear radii of two atoms is 2:3 then the ratio of their mass numbers is: (A) 8:27 (B) 4:9 (C) 27:8 (D) 9:4
›Reveal solutionSolution
The ratio of nuclear radii is given by R∝A1/3, so a radius ratio of 2:3 implies a mass-number ratio of (2)3:(3)3=8:27. The correct option is (A).
The key idea here is that nuclear radius scales with the cube root of the mass number. This comes from the fact that nuclei are roughly spherical and have nearly constant density — so volume (and thus radius cubed) is proportional to the number of nucleons.
- Recall the nuclear radius formula For a nucleus, the radius is approximately
R=R0A1/3
where R0 is a constant (about 1.2×10−15 m) and A is the mass number. This relation holds because nuclear matter is incompressible: adding nucleons increases volume proportionally.
- Set up the ratio If two nuclei have radii R1 and R2 with ratio
R2R1=32,
then using the formula:
R2R1=R0A21/3R0A11/3=(A2A1)1/3.
- Solve for the mass-number ratio Equate the two expressions:
(A2A1)1/3=32.
Cube both sides:
A2A1=(32)3=278.
- Interpret the result …
- COMEDK 2025Set 2025-E1 markMCQQ.If the mass numbers of two nuclei are in the ratio 5:2 and their diameters are in ratio 2:6. Then their nuclear densities will be in the ratio (A) 1:1 (B) 2:5 (C) 10:12 (D) 6:5
›Reveal solutionSolution
Nuclear density is independent of the size of the nucleus — it is constant for all nuclei. Therefore, the ratio of nuclear densities is always 1:1, regardless of given mass or diameter ratios.
Concept & Intuition
The key idea is that nuclear density is roughly constant for all atoms. Why? Because the volume of a nucleus is proportional to its mass number A (the number of nucleons). Since mass is also proportional to A, the density ρ=volumemass becomes independent of A. This is a fundamental result of the liquid-drop model of the nucleus: nucleons pack together at a fixed density, like drops of an incompressible fluid.
Step-by-step reasoning
- Recall the empirical formula for nuclear radius The radius R of a nucleus is given by
R=R0A1/3
where R0≈1.2×10−15m is a constant, and A is the mass number.
- Express the volume in terms of A The volume V of a spherical nucleus is
V=34πR3=34π(R0A1/3)3=34πR03A
So V∝A.
-
Relate mass to mass number
The mass of a nucleus is approximately m≈A⋅u, where u is the atomic mass unit. Hence m∝A.
-
Compute nuclear density
ρ=Vm∝AA=constant
Therefore, nuclear density is the same for all nuclei.
- Apply to the given ratios …
- COMEDK 2024Set 2024-A1 markMCQQ.The radius of a nucleus as measured by electron scattering is 4.8 fm. The mass number of nucleus is most likely to be (A) 46 (B) 16 (C) 64 (D) 48
›Reveal solutionSolution
From R=R0A1/3 with R0=1.2 fm, a radius of 4.8 fm gives A1/3=4, so A=64.
The nuclear radius follows R=R0A1/3 with R0≈1.2 fm. Solve for A: …
- COMEDK 2024Set 2024-E1 markMCQQ.
[!FORMULA] If the nuclear radius of 27Al is 3.6 fermi, the nuclear radius of 125Fe is
(A) 6×10−10 m (B) 6×10−13 m (C) 6×10−15 m (D) 6×10−12 m›Reveal solutionSolution
The nuclear radius scales as R=R0A1/3, so the ratio of radii is the cube root of the mass-number ratio. For 27Al (A=27, R=3.6 fm) and 125Fe (A=125), the radius is 6.0 fm = 6×10−15 m, which matches option (C).
The key concept is the empirical nuclear radius formula: R=R0A1/3, where R0 is a constant (≈ 1.2 fm) and A is the mass number. This arises because nuclear matter has roughly constant density — like a liquid drop — so volume is proportional to A, and radius is proportional to A1/3.
Why this works:
If you double the number of nucleons, you double the volume, so the radius increases by a factor of 21/3≈1.26. Here we know the radius for A=27 and want it for A=125. Instead of finding R0, we can directly use the ratio.
-
Write the radius formula for both nuclei:
For 27Al: RAl=R0(27)1/3
For 125Fe: RFe=R0(125)1/3
-
Take the ratio to eliminate R0:
RAlRFe=(27)1/3(125)1/3=(27125)1/3
- Simplify the cube roots: 125=53 and 27=33, so
(27125)1/3=35
- Plug in the known radius: …
-
- COMEDK 2024Set 2024-M1 markMCQQ.The ratio of the radii of the nucleus of two element X and Y having the mass numbers 232 and 29 is: (A) 4 : 1 (B) 1 : 4 (C) 1 : 2 (D) 2 : 1
›Reveal solutionSolution
The nuclear radius scales as the cube root of the mass number, so the ratio of radii for mass numbers 232 and 29 is found by taking the cube root of their ratio, giving 2:1.
The key concept here is the nuclear radius formula. In nuclear physics, the radius of a nucleus is not proportional to its mass number A directly, but rather to A1/3. This is because the nucleus is roughly spherical and its volume is proportional to the number of nucleons (protons and neutrons). Since volume scales as r3, the radius scales as A1/3. The standard empirical formula is:
R=R0A1/3
where R0 is a constant (about 1.2×10−15 m). For comparing two nuclei, the constant cancels out.
- Write the radii for both elements. For element X with mass number AX=232:
RX=R0(232)1/3
For element Y with mass number AY=29:
RY=R0(29)1/3
- Take the ratio of the radii. The constant R0 cancels, so:
RYRX=(29)1/3(232)1/3=(29232)1/3
- Simplify the fraction inside the cube root.
29232=8
(Since 29×8=232.)
- Take the cube root.
- COMEDK 2023Set 2023-E1 markMCQQ.The mass number of two nuclei P and Q are 27 and 125 respectively. The ratio of their radii RP:RQ is given by: (A) 9 : 25 (B) 3 : 5 (C) 27 : 25 (D) 5 : 3
›Reveal solutionSolution
R_P / R_Q = (A_P / A_Q)^(1/3) = (27 / 125)^(1/3) = 3/5.
Concept: nuclear radius R = R0 * A^(1/3), where A is the mass number (nuclear density is essentially constant). …
- COMEDK 2023Set 2023-M1 markMCQQ.The mass density of a nucleus varies with mass number A as (A) Ao (B) A2 (C) A1 (D) lnA
›Reveal solutionSolution
Because the nuclear radius scales as A1/3, both mass and volume grow linearly in A, so nuclear density is constant — independent of A (A0).
Nuclear radius: R=R0A1/3, so volume V=34πR3=34πR03A∝A.
Mass ≈AmN∝A. Therefore density: …
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