The radius of atom is of the order of 1 Å and radius of nucleus is of the order of fermi. How many magnitudes

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Q. The radius of atom is of the order of 1 Å and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

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Concept understanding — Atomic Volume Calculation

Atomic Volume: Meaning and Calculation

Atoms are mostly empty space — a tiny dense nucleus wrapped in a fuzzy electron cloud. So "atomic volume" does not mean the volume of a solid ball; it means the average space one atom occupies when many atoms are packed together in a solid or liquid.

Think of a crowded hall: to find the space per person you divide the hall's volume by the number of people. Atomic volume does exactly that for atoms.

Definition

Atomic volume is the volume occupied by one mole of atoms of an element in its solid or liquid state. It is found from the element's molar mass and density:

V_{\text{atomic}} = \frac{M}{\rho}

$V_{\text{atomic}}$ = atomic volume (cm³/mol)
$M$ = molar mass (g/mol)
$\rho$ = density (g/cm³)

This gives the volume per mole. Dividing by Avogadro's number gives the space per single atom:

V_{\text{one atom}} = \frac{V_{\text{atomic}}}{N_A}, \qquad N_A = 6.022\times10^{23}\ \text{mol}^{-1}

Worked example — aluminium

Molar mass $M = 26.98$ g/mol, density $\rho = 2.70$ g/cm³:

V_{\text{atomic}} = \frac{26.98}{2.70} = 9.99\ \text{cm}^3/\text{mol}

So one mole of Al atoms occupies about 10 cm³. Per atom:

V_{\text{one Al atom}} = \frac{9.99}{6.022\times10^{23}} = 1.66\times10^{-23}\ \text{cm}^3

Periodic trends

Down a group: atomic volume increases — more electron shells make atoms larger.
Across a period: it generally decreases — rising nuclear charge pulls the electrons in tighter.
Allotropes differ: diamond is denser than graphite, so diamond's atomic volume is smaller, though both are carbon.

These trends help explain why alkali metals (large atomic volume) are soft and reactive, while transition metals (smaller atomic volume) are hard and dense. …

Why this formula?

Atomic Volume Calculation: Understanding the "Why" Behind the Formula

What Is Atomic Volume?

Atomic volume is not the volume of a single atom — it's the volume occupied by one mole of atoms of an element in its solid state. This is a macroscopic quantity that helps us understand how tightly atoms pack together.

The key formula is:

\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}

Let's break down why this works.

The Core Reasoning: From Mass to Volume

Step 1: What does density tell us?

Density ( $\rho$ ) is defined as:

\rho = \frac{\text{Mass}}{\text{Volume}}

For a pure solid element, if we take one mole of atoms:

Mass of one mole = Atomic mass (in g/mol)
Volume of one mole = Atomic volume (in cm³/mol)

So:

\rho = \frac{\text{Atomic mass}}{\text{Atomic volume}}

Step 2: Rearranging to find atomic volume

\text{Volume} = \frac{\text{Mass}}{\text{Density}}

Therefore:

\boxed{\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}}

Why This Makes Physical Sense

Atomic mass tells you how heavy one mole of atoms is; density tells you how much mass fits in a given space. Dividing mass by density gives the space that mass occupies.

Example intuition: If iron has atomic mass ≈ 56 g/mol and density 7.87 g/cm³, then:

\text{Atomic volume} = \frac{56}{7.87} \approx 7.1 \text{ cm}^3/\text{mol}

This means one mole of iron atoms (about $6.022 \times 10^{23}$ atoms) occupies roughly 7.1 cm³ of space.

Important Exam Points

Concept	Why It Matters
Units	Atomic mass in g/mol, density in g/cm³ → atomic volume in cm³/mol
Solid state only	The formula assumes atoms are closely packed; gases/liquids have different packing

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The volume of an atom is about $10^{15}$ times larger than the volume of its nucleus. This enormous ratio comes from the cubic dependence of volume on radius — since the atomic radius ( $1 \text{ Å} = 10^{-10} \text{ m}$ ) is $10^5$ times the nuclear radius ( $1 \text{ fermi} = 10^{-15} \text{ m}$ ), the volume ratio is $(10^5)^3 = 10^{15}$ .

The key idea here is simple but profound: volume scales as the cube of the radius. So even a modest difference in radius leads to a dramatic difference in volume. Let’s see why that matters for an atom and its nucleus.

An atom is mostly empty space. Its mass is concentrated in the tiny nucleus at the centre, while the electron cloud defines the atom’s overall size. The typical atomic radius is about $1 \text{ Å} = 10^{-10} \text{ m}$ . The nuclear radius is roughly $1 \text{ fermi} = 10^{-15} \text{ m}$ . That’s a factor of $10^5$ in linear size. But volume? That’s where the real gap appears.

Write down the radii in consistent units.

Atomic radius: $R_a = 1 \text{ Å} = 10^{-10} \text{ m}$ .

Nuclear radius: $R_n = 1 \text{ fermi} = 10^{-15} \text{ m}$ .
Assume both are spheres.

The volume of a sphere is $V = \frac{4}{3}\pi R^3$ . So the ratio of volumes is:

\frac{V_a}{V_n} = \frac{\frac{4}{3}\pi R_a^3}{\frac{4}{3}\pi R_n^3} = \left(\frac{R_a}{R_n}\right)^3.

Plug in the numbers.

\frac{R_a}{R_n} = \frac{10^{-10}}{10^{-15}} = 10^{5}.

Therefore:

$\frac{V_a}{V_n} = (10^{5})^3 = 10^{15}.$ …

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Common Mistakes in Atomic Volume Calculation

Students often slip on this problem because of unit handling and ratio logic. Here are the most frequent errors — and how to avoid each.

✗ Mistake 1: Forgetting to cube the radius ratio

What students do wrong:

They compare radii directly:

$\frac{r_{\text{atom}}}{r_{\text{nucleus}}} = \frac{10^{-10}}{10^{-15}} = 10^5$

Then they say the volume is also $10^5$ times larger.

Why it’s wrong:

Volume scales with $r^3$ , not $r$ .

✓ How to avoid:

Always write the volume ratio formula first:

\frac{V_{\text{atom}}}{V_{\text{nucleus}}} = \frac{\frac{4}{3}\pi r_{\text{atom}}^3}{\frac{4}{3}\pi r_{\text{nucleus}}^3} = \left( \frac{r_{\text{atom}}}{r_{\text{nucleus}}} \right)^3

Then substitute:

\left( \frac{10^{-10}}{10^{-15}} \right)^3 = (10^5)^3 = 10^{15}

Key result: The atom’s volume is $10^{15}$ times larger than the nucleus’s volume.

✗ Mistake 2: Incorrect unit conversion (Å to m)

What students do wrong:

They treat 1 Å as $10^{-10}$ m correctly, but then use 1 fermi = $10^{-13}$ cm or $10^{-15}$ m inconsistently.

✓ How to avoid:

Memorise the standard conversions:

1 Å = $10^{-10}$ m
1 fermi (fm) = $10^{-15}$ m

Always convert both to metres before comparing. Never mix Å and fm directly without converting.

✗ Mistake 3: Confusing “order of magnitude” with exact value

What students do wrong:

They try to compute exact numbers like $1.2 \times 10^{-10}$ m and $1.4 \times 10^{-15}$ m, then cube them — leading to messy arithmetic and errors.

✓ How to avoid:

The problem says “of the order of” — so use powers of 10 only:

$r_{\text{atom}} \approx 10^{-10}$ m
$r_{\text{nucleus}} \approx 10^{-15}$ m

Then:

\left( \frac{10^{-10}}{10^{-15}} \right)^3 = 10^{15}

No calculator needed. The answer is 15 orders of magnitude.

✗ Mistake 4: Forgetting to state “orders of magnitude”

What students do wrong: …

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←Previous questionQ19 — Why do we have different units for the same physical quantity?→Next questionQ21 — Name the device used for measuring the mass of atoms and molecules.

Q.The radius of atom is of the order of 1 Å and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

Concept understanding — Atomic Volume Calculation

Atomic Volume: Meaning and Calculation

Definition

Worked example — aluminium

Periodic trends

Why this formula?

Atomic Volume Calculation: Understanding the "Why" Behind the Formula

What Is Atomic Volume?

The Core Reasoning: From Mass to Volume

Step 1: What does density tell us?

Step 2: Rearranging to find atomic volume

Why This Makes Physical Sense

Important Exam Points

Method: Direct Volume Ratio Using Spherical Approximation

Steps

Common Mistakes in Atomic Volume Calculation

✗ Mistake 1: Forgetting to cube the radius ratio

✗ Mistake 2: Incorrect unit conversion (Å to m)

✗ Mistake 3: Confusing “order of magnitude” with exact value

✗ Mistake 4: Forgetting to state “orders of magnitude”

Q.The radius of atom is of the order of 1 Å and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

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