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. …
- CBSE 2026Set A1 markMCQQ.The nuclear density is approximately (A) independent of mass number (B) directly proportional to mass number (C) inversely proportional to mass number (D) directly proportional to A^(1/3)
›Reveal solutionSolution
Nuclear radius R = R₀A^(1/3), so volume ∝ A and mass ∝ A, making density independent of A.
The nuclear radius follows the empirical relation R=R0A1/3, where R0≈1.2 fm and A is the mass number.
Volume of the nucleus: V=34πR3=34πR03A, so V∝A.
Mass of the nucleus: M≈A×mnucleon, so M∝A.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be(a) (3)^(1/3) : 1(b) 1:1(c) 1:3(d) 3:1
›Reveal solutionSolution
Because the nuclear radius scales as R = R0 A^(1/3), the nuclear volume scales exactly as A, so density = mass/volume comes out essentially the same constant for every nucleus, regardless of A.
Nuclear radius: R = R0 A^(1/3), where R0 approx 1.2 fm is a constant.
Nuclear volume: V = (4/3) pi R^3 = (4/3) pi R0^3 A
Since nuclear mass is approximately proportional to A (mass number, roughly A times the nucleon mass), density:
rho = mass / volume is proportional to A / A = constant
…
- CBSE 2025Set ANNUAL1 markQ.The density of nuclear matter is independent of the size of the nucleus. (T/F)
›Reveal solutionSolution
This statement is True — nuclear density is essentially the same for all nuclei, independent of the mass number (size) of the nucleus.
The radius of a nucleus with mass number A is given empirically by R=R0A1/3, where R0≈1.2 fm is a constant. The nuclear volume is then V∝R3∝A, and since the nuclear mass is also proportional to A (each nucleon has roughly the same mass), the density …
- CBSE 2024Set ANNUAL1 markQ.What is the ratio of nuclear densities of two nuclei having mass number 1:4?
›Reveal solutionSolution
Nuclear density is (almost exactly) the same for all nuclei, regardless of mass number.
Nuclear density is given by ρ=volumemass=34πR3mA, where the nuclear radius scales as R=R0A1/3. Substituting:
ρ=34π(R0A1/3)3mA=34πR03AmA=34πR03m
…
- CBSE 2024Set ANNUAL1 markMCQQ.Two nuclei have mass numbers in the ratio 1:3, the ratio of the nuclear densities are(a) 3:1(b) 1:1(c) 1:9(d) 1:3
›Reveal solutionSolution
Nuclear density ρ=mass/volume∝A/A3⋅3= constant, since the nuclear radius R∝A1/3 makes volume ∝A — so it does not depend on mass number at all.
The empirical nuclear radius formula is R=R0A1/3, where R0≈1.2fm is a constant and A is the mass number. The nuclear volume is
V=34πR3=34πR03A
which is directly proportional to A. The nuclear mass is also (to good approximation) proportional to A, since M≈Au (each nucleon contributes roughly one atomic mass unit).
So the nuclear density is …
- CBSE 2023Set 55/1/11 markMCQQ.The mass density of a nucleus of mass number A is :(a) proportional to A1/3(b) proportional to A2/3(c) proportional to A3(d) independent of A
›Reveal solutionSolution
The mass density of a nucleus is roughly constant for all nuclei because both the mass and the volume scale with the mass number A, making the density independent of A. The correct option is (d).
The key idea here is that a nucleus behaves like a tiny, incompressible drop of nuclear matter. Its volume is proportional to the number of nucleons (protons and neutrons), and its mass is also proportional to that number. When you take the ratio, the A cancels out.
Let’s see why this is true step by step.
-
What is mass number A?
A is the total number of nucleons in the nucleus. The mass of a single nucleon is roughly mn≈1.67×10−27 kg. So the mass of the nucleus is approximately M≈A⋅mn. This is a direct proportionality: M∝A.
-
How does the size of a nucleus scale with A?
Experiments (like Rutherford scattering) show that the radius R of a nucleus follows the empirical formula:
R=R0A1/3
where R0≈1.2×10−15 m (about 1.2 femtometers). This is a well-established result — the nuclear volume grows with the number of nucleons.
- What is the volume of the nucleus? Treating the nucleus as a sphere:
V=34πR3=34π(R0A1/3)3=34πR03A
So V∝A. The volume is directly proportional to the number of nucleons.
- Now compute the density: …
-
- CBSE 2022Set ANNUAL1 markQ.Two nuclei have mass numbers in the ratio 1 : 27. What is the ratio of nuclear density?
›Reveal solutionSolution
Nuclear density is the same for all nuclei, so the ratio is 1 : 1.
The nuclear radius is R=R0A1/3, so the nuclear volume is
V=34πR3=34πR03A.
The nuclear mass is proportional to the mass number, m≈Amnucleon. Hence the density is
ρ=Vm=34πR03AAmnucleon=34πR03mnucleon, …
- CBSE 2020Set 55/1/11 markQ.The nuclear radius of 1327Al is 3⋅6 fermi. Find the nuclear radius of 2964Cu.(OR)A proton and an electron have equal speeds. Find the ratio of de Broglie wavelengths associated with them.
›Reveal solutionSolution
Part (a): Since R∝A1/3, RCu=RAl(64/27)1/3=3.6×34=4.8 fermi.
Part (b): With equal speeds λ∝1/m, so λp/λe=me/mp≈1/1836.
Part (a)
The nucleus behaves like a constant-density drop, so its volume ∝A and its radius follows
R=R0A1/3,R0≈1.2 fm.
Taking a ratio removes R0:
RAlRCu=(AAlACu)1/3=(2764)1/3=34.
With RAl=3.6 fm:
RCu=3.6×34=4.8 fm. …
- CBSE 2020Set ANNUAL1 markMCQQ.The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as :(a) A5/3(b) A2/3(c) A4/3(d) A1/3
›Reveal solutionSolution
Since the nuclear radius scales as A1/3, the surface area of the (spherical) nucleus scales as (A1/3)2=A2/3.
Working
The empirical nuclear radius formula is
R=R0A1/3,R0≈1.2 fm
Treating the nucleus as a sphere, its surface area is …
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