Q.Given the mass of iron nucleus as 55.85 u and A=56, find the nuclear density.
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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 density is nearly constant for all nuclei because the nuclear volume scales linearly with mass number A.
Step 1 – Find the nuclear radius.
Using the empirical formula R=R0A1/3, where R0=1.2×10−15 m:
R=1.2×10−15×(56)1/3 m.
Since 561/3≈3.83,
R≈4.60×10−15 m.
Step 2 – Compute the nuclear volume.
V=34πR3=34π(4.60×10−15)3≈4.07×10−43 m3.
Step 3 – Convert mass to kg and find density. …
Nuclear density is nearly constant for all nuclei because the nuclear volume scales linearly with mass number A. Using the iron nucleus (A=56, mass =55.85 u) and the empirical radius formula R=R0A1/3 with R0=1.2 fm, the density comes out to about 2.3×1017 kg/m3.
The idea behind nuclear density is beautiful in its simplicity. Unlike ordinary matter, where density varies wildly from gas to solid, nuclear matter has an almost constant density. Why? Because a nucleus is a tightly packed sphere of protons and neutrons. If you add more nucleons, the volume increases proportionally — the radius follows R=R0A1/3, so volume ∝A. Mass also ∝A (since each nucleon has roughly 1 u). So density ≈ constant, independent of A.
We’ll now calculate it for iron, step by step.
- Convert the mass to kilograms. The mass of the iron nucleus is given as 55.85 u. One atomic mass unit is 1 u=1.660539×10−27 kg. So:
m=55.85×1.660539×10−27 kg≈9.27×10−26 kg.
- Find the nuclear radius. The empirical formula for nuclear radius is:
R=R0A1/3,
where R0≈1.2 fm (1 femtometre = 10−15 m).
For iron, A=56, so:
R=1.2×10−15×561/3 m.
Now 561/3 is about 3.825 (since 3.83=54.9, close enough).
Thus:
R≈1.2×10−15×3.825≈4.59×10−15 m.
- Compute the volume. The nucleus is spherical, so:
V=34πR3.
First cube the radius:
R3≈(4.59×10−15)3=4.593×10−45≈96.7×10−45=9.67×10−44 m3.
Then:
V=34π×9.67×10−44≈4.1888×9.67×10−44≈4.05×10−43 m3.
- Calculate density. Density ρ=Vm: …
Method: Direct Application of the Nuclear Density Formula
This problem uses the fact that nuclear density is nearly constant for all nuclei. The method is straightforward: find the nuclear volume from the radius formula, then divide mass by volume.
Step 1: Write the nuclear radius formula
The radius of a nucleus is given by:
R=R0A1/3
where R0=1.2×10−15 m (a constant) and A is the mass number.
Step 2: Convert the given mass to kilograms
Mass of iron nucleus = 55.85 u.
Recall: 1 u=1.66×10−27 kg.
So:
m=55.85×1.66×10−27=9.27×10−26 kg
The mass number A=56 is close to the mass in u (55.85). This is because 1 u ≈ mass of one nucleon. For density calculations, using either value gives nearly the same result.
Step 3: Calculate the nuclear radius
R=(1.2×10−15)×(56)1/3
561/3≈3.83 (since 3.833=56.2).
R=1.2×10−15×3.83=4.60×10−15 m
Step 4: Calculate the nuclear volume
The nucleus is spherical:
V=34πR3
First find R3:
R3=(4.60×10−15)3=97.3×10−45=9.73×10−44 m3
Then:
V=34×3.14×9.73×10−44 …
The most common mistake here is treating the mass number A as the mass of the nucleus in kilograms. A is just the number of nucleons — it has no units. The mass in kilograms must be calculated separately.
Mistake 1: Using A directly as mass in kg
A student writes ρ=34πR356 and gets a wildly wrong answer. The mass number 56 is dimensionless, not a mass. You must convert the given mass from atomic mass units (u) to kg first: 1 u=1.66×10−27 kg.
Never plug A into the density formula as if it were the mass. A only tells you the number of nucleons, not the mass in SI units.
Mistake 2: Forgetting the nuclear radius formula
The radius of a nucleus is R=R0A1/3, where R0≈1.2×10−15 m. Some students use the atomic radius (of the order 10−10 m) instead, which makes the density off by a factor of 1015. The nucleus is tiny — always use the nuclear radius formula.
Mistake 3: Using the atomic mass instead of nuclear mass
The problem gives the mass of the iron nucleus as 55.85 u, so this is already correct. But if a question gives the atomic mass, remember that the mass of electrons is included. For iron (Z=26), that’s about 26×9.1×10−31 kg — negligible for most exam purposes, but conceptually you should know the difference.
Mistake 4: Unit mismatch in the final answer
Nuclear density comes out around 2.3×1017 kg/m3. A common slip is reporting it in g/cm3 without converting, or forgetting that 1 u=1.66×10−27 kg and 1 fm=10−15 m.
Work entirely in SI units (kg, m) from the start. Convert u to kg and fm to m before plugging into any formula. This avoids unit errors at the end. …
Showing the 12 most recent of 13 on this concept.
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.If the radius of a nucleus having 13 protons and 14 neutrons is 3.6 fm, then the ratio of the volume to surface area of nucleus having 53 protons and 72 neutrons is (A) 12 fm (B) 3 fm (C) 6 fm (D) 2 fm
›Reveal solutionSolution
Uses the empirical nuclear-radius formula R=R0A1/3 to find the radius of the second nucleus, then the simple geometric fact that a sphere's volume-to-surface-area ratio is R/3.
Concept and Intuition
Nuclei are modelled as spheres whose radius grows with mass number as R=R0A1/3, reflecting the (nearly) constant nuclear density (constant volume per nucleon). Once we know R0 from one nucleus, we can predict the radius of any other nucleus purely from its mass number. The ratio of volume to surface area for any sphere is a purely geometric fact: SV=4πR234πR3=3R.
Step-by-Step Solution
- First nucleus: A1=13+14=27 nucleons, R1=3.6 fm. Since 271/3=3: R0=R1/A11/3=3.6/3=1.2 fm. …
- AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQQ.If the nuclear radius of 27Al is 3.6 fermi, the approximate nuclear radius of 64Cu in fermi is (A) 2.4 (B) 1.2 (C) 4.8 (D) 3.6
›Reveal solutionSolution
Using R=R0A1/3 and the given Al radius to find R0, the Cu-64 nuclear radius comes out to 4.8 fm.
Concept and Intuition
Nuclear radius follows the empirical law R=R0A1/3, reflecting the fact that nuclear matter has roughly constant density — so volume (and hence R3) scales directly with the number of nucleons A. Given one nucleus's radius, we can extract the constant R0 and then predict any other nucleus's radius from its mass number.
Step-by-Step Solution
- For 27Al: RAl=R0(27)1/3=R0×3=3.6fm⇒R0=1.2fm. …
- AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQQ.The ratio of the radii of the nuclei of mass numbers 27 and 64 is (A) 3 : 4 (B) 4 : 3 (C) 9 : 16 (D) 16 : 9
›Reveal solutionSolution
Nuclear radius scales as the cube root of mass number; recognizing 27 and 64 as perfect cubes gives a clean 3:4 ratio.
Concept and Intuition
Since nuclear density is roughly constant across nuclei, the volume (and hence R3) is proportional to the mass number A, giving R=R0A1/3.
Step-by-Step Solution
- R1/R2=(A1/A2)1/3=(27/64)1/3.
- 27=33 and 64=43, so (27/64)1/3=3/4. …
- AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQQ.The density (in kgm−3) of nuclear matter is of the order of (A) 1021 (B) 1017 (C) 1012 (D) 108
›Reveal solutionSolution
Nuclear density is a constant, extremely large number (independent of which nucleus), of order 1017kg m−3.
Concept and Intuition
Experiments show the nuclear radius follows R=R0A1/3 with R0≈1.2×10−15m, where A is the mass number. Since volume ∝R3∝A, and mass ∝A (each nucleon has roughly the same mass), the density ρ=mass/volume comes out independent of A — every nucleus, light or heavy, has almost the same density. This is a striking fact used to show nuclear matter is incompressible and nucleons are tightly packed.
Step-by-Step Solution
- Take a representative nucleus, e.g., with mass number A, radius R=1.2×10−15A1/3m.
- Mass ≈A×1.67×10−27kg (mass of one nucleon).
- Volume =34πR3=34π(1.2×10−15)3A.
- Compute (1.2×10−15)3≈1.73×10−45m3, so volume ≈34π×1.73×10−45A≈7.24×10−45Am3. …
- AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQQ.If the surface areas of two nucleii are in the ratio 9:49, then the ratio of their mass numbers is (A) 27:343 (B) 9:49 (C) 3:7 (D) 49:81
›Reveal solutionSolution
Since nuclear radius goes as A1/3, surface area goes as A2/3; inverting the given area ratio gives a mass-number ratio of 27:343.
Concept and Intuition
Nuclei are modeled as spheres with radius R=R0A1/3, where A is the mass number — this comes from nuclear density being roughly constant (volume ∝A, and volume ∝R3, so R∝A1/3). Surface area of a sphere is 4πR2, so area ∝A2/3.
Step-by-Step Solution
- Write the area ratio in terms of mass numbers: S2S1=(A2A1)2/3=499.
- Solve for the mass-number ratio by raising both sides to the power 3/2: A2A1=(499)3/2. …
- AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQQ.A nucleus with atomic mass number 'A' produces another nucleus by loosing 2 alpha particles. The volume of the new nucleus is 60 times that of the alpha particle. The atomic mass number A of the original nucleus is (A) 228 (B) 238 (C) 248 (D) 244
›Reveal solutionSolution
Nuclear volume is proportional to mass number (via R=R0A1/3); losing two alpha particles
removes 8 from A, and matching the volume ratio (60 times an alpha's volume) gives A = 248.
Concept and Intuition
Since nuclear radius R=R0A1/3, nuclear volume V∝R3∝A — directly
proportional to the mass number, with the same constant of proportionality for any nucleus
(including an alpha particle, which has mass number 4).
Step-by-Step Solution
- Emitting 2 alpha particles removes a total mass number of 2×4=8 from the original nucleus (mass number A), leaving a daughter nucleus of mass number A−8.
- Since volume ∝ mass number, the ratio of the daughter's volume to an alpha particle's …
- AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQQ.The surface areas of two nucleii are in the ratio 9:25. The mass numbers of the nucleii are in the ratio (A) 27:125 (B) 9:25 (C) 3:5 (D) 1:1
›Reveal solutionSolution
Nuclear radius scales as R∝A1/3, so surface area scales as A2/3; invert to find the mass-number ratio. Answer: 27:125.
Concept and Intuition
The nuclear radius formula R=R0A1/3 tells us nuclear volume scales with mass number A (nucleons packed at roughly constant density), and hence radius scales as A1/3. Surface area, being proportional to R2, then scales as A2/3.
Step-by-Step Solution
- Surface area ratio =R12:R22=9:25, so R1:R2=3:5.
- Since R∝A1/3, A11/3:A21/3=3:5.
- Cubing both sides: A1:A2=33:53=27:125. …
- AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQQ.The ratio of the radii of the nuclei 29X64 and 84Y216 is (A) 32 (B) 158 (C) 38 (D) 87
›Reveal solutionSolution
Nuclear radius scales as A1/3; taking the cube root of the mass-number ratio 64/216 gives 2/3.
Concept and Intuition
Since nucleons are packed at roughly constant density inside a nucleus, the nuclear volume ∝A (mass number), so the radius R∝A1/3. This is why the atomic number (Z) is irrelevant to this ratio — only the mass number matters.
Step-by-Step Solution
- R=R0A1/3, so RYRX=(AYAX)1/3=(21664)1/3.
- 64=43 and 216=63. …
- AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQQ.The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is (A) 64 (B) 81 (C) 100 (D) 125
›Reveal solutionSolution
Nuclear radius scales as A1/3; use the given (A, R) pair to fix R0, then solve for the mass number at R=6 fm, giving A=125.
Concept and Intuition
The empirical nuclear radius formula R=R0A1/3 reflects that nucleons pack at roughly constant density, so nuclear volume (and hence R3) is proportional to the number of nucleons A.
Step-by-Step Solution
- R=R0A1/3. For A=64: 641/3=4, so 4.8=R0×4⇒R0=1.2 fm.
- For the new nucleus, 6=1.2A1/3⇒A1/3=5. …
- AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQQ.Choose the correct statement of the following (A) The nuclear density in general, is independent of mass number A. (B) The radius of a nucleus is directly proportional to the mass number A of the nucleus. (C) The binding energy of a nucleus is inversely proportional to its mass defect. (D) Energy is absorbed when heavy nuclei undergo transmutation into light nuclei.
›Reveal solutionSolution
Nuclear density is essentially constant across all nuclei because both mass and volume scale the same way with mass number A — this is the key experimental fact underlying the liquid-drop model.
Concept and Intuition
Nuclear radius empirically follows R=R0A1/3 (nucleons are packed at essentially constant density, like an incompressible fluid drop). Since nuclear mass m≈Amp scales as A, and volume V=34πR3∝(A1/3)3=A, the density ρ=m/V∝A/A= constant, independent of A.
Step-by-Step Solution
- Nuclear radius: R=R0A1/3, so volume V∝R3∝A.
- Nuclear mass ∝A (roughly A nucleons of similar mass each).
- Density =volumemass∝AA= constant, i.e. independent of A — confirms (A).
- (B) is wrong: radius ∝A1/3, not ∝A.
- (C) is wrong: binding energy BE=(Δm)c2 is directly proportional to mass defect Δm, not inversely. …
- AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQQ.A nucleus has mass number A1 and volume V1. Another nucleus has mass number A2 and volume V2. If the relation between the mass numbers is A2=3A1, then V2V1= (A) 331 (B) (31)31 (C) 31 (D) 31
›Reveal solutionSolution
Nuclear volume is directly proportional to mass number (V∝A since R∝A1/3), so V1/V2=A1/A2=1/3.
Concept and Intuition
A key nuclear physics fact is that nuclear density is (approximately) constant across nuclei — the radius scales as R∝A1/3 so that volume V∝R3∝A. This directly implies that a nucleus with 3 times the mass number has 3 times the volume, not 3 times the radius.
Step-by-Step Solution
- Nuclear radius formula: R=R0A1/3.
- Nuclear volume: V=34πR3=34πR03A, i.e. V∝A. …
- AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQQ.Ratio of density of nuclear matter to density of water is at least ____ (R0=1.2×10−15m & mp=mn=1.67×10−27kg) (A) 2.307×1014kg.m−3 (B) 2.307×1017kg.m−3 (C) 23.07×1014kg.m−3 (D) 23.07×1017kg.m−3
›Reveal solutionSolution
Nuclear density comes out to ρnucleus≈2.307×1017kgm−3 regardless of mass number; dividing by water's density (103kgm−3) gives the ratio 2.307×1014.
Concept and Intuition
Because the nuclear radius scales as R=R0A1/3, the nuclear volume scales exactly as A (mass number), which cancels the A in the numerator when computing density — nuclear density is essentially the same for every nucleus, a striking and often-tested fact.
Step-by-Step Solution
- Mass of nucleus ≈Amp (taking mp≈mn).
- Volume =34πR3=34π(R0A1/3)3=34πR03A.
- Density ρ=34πR03AAmp=34πR03mp — the A cancels.
- Compute R03=(1.2×10−15)3=1.728×10−45m3.
- 34πR03≈4.189×1.728×10−45≈7.238×10−45m3. …
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