The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $10^7\ \text{K}$, and its outer surface at a temperature of about $6000\ \text{K}$. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun $= 2.0 \times 10^{30}\ \text{kg}$, radius of the Sun $= 7.0 \times 10^{8}\ \text{m}$.

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷K, and its outer su

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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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Dividing the Sun's mass by its volume gives an average density of about $1.4\times10^3\ \text{kg/m}^3$ — comparable to water, in the range of liquids and solids, not gases, even though the Sun is entirely plasma.

Why "hot plasma" doesn't automatically mean "low density"

It's tempting to guess that because the Sun is a superheated, ionized gas, its density should resemble ordinary gases (of order $1\ \text{kg/m}^3$ ). But density depends on pressure, not just temperature — and the core of the Sun is compressed by its own enormous gravity to a pressure of hundreds of billions of atmospheres. Let's check the guess against the actual numbers.

Computing the average density

\rho = \frac{M}{V}, \qquad V = \frac{4}{3}\pi R^3

Given: $M = 2.0\times10^{30}\ \text{kg}$ , $R = 7.0\times10^{8}\ \text{m}$ .

Volume:

R^3 = (7.0\times10^{8})^3 = 343\times10^{24} = 3.43\times10^{26}\ \text{m}^3

V = \frac{4}{3}\pi \times 3.43\times10^{26} \approx 1.44\times10^{27}\ \text{m}^3

Density:

\rho = \frac{2.0\times10^{30}}{1.44\times10^{27}} \approx 1.4\times10^{3}\ \text{kg/m}^3

Interpreting the result …

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Common Mistakes in Atomic Volume / Density Calculation (Sun Problem)

Students often make predictable errors when solving this NCERT/competitive exam problem. Here are the most frequent ones and how to avoid each:

1. ✗ Using wrong formula for volume of a sphere

Mistake: Using $V = \pi r^2$ (area) or $V = \frac{4}{3}\pi r^2$ (missing one power of $r$ ).

How to avoid: Memorise the exact formula:

V = \frac{4}{3}\pi r^3

Always write it down before substituting numbers. Double-check the exponent on $r$ .

2. ✗ Forgetting to cube the radius correctly

Mistake: Computing $r^3$ as $7.0 \times 10^{24}$ instead of $7.0^3 \times 10^{24} = 343 \times 10^{24}$ .

How to avoid:

Cube the coefficient and the power of 10 separately:

(7.0 \times 10^8)^3 = 7.0^3 \times (10^8)^3 = 343 \times 10^{24}

Write intermediate steps — don’t skip directly to the final number.

3. ✗ Mixing up units (km vs m)

Mistake: Using radius in km ( $7.0 \times 10^5$ km) but mass in kg, leading to density off by $10^9$ .

How to avoid:

Always convert to SI units before calculation.
Given radius is already in metres ( $7.0 \times 10^8$ m) — confirm this before proceeding.

4. ✗ Arithmetic errors in powers of 10

Mistake: Getting $10^{30} / 10^{24} = 10^6$ instead of $10^{30} / 10^{25} = 10^5$ (if volume exponent is wrong).

How to avoid:

Compute volume exponent carefully:

$r^3 \to (10^8)^3 = 10^{24}$ , then multiply by $\frac{4}{3}\pi \approx 4.19$ → exponent stays $10^{24}$ (coefficient ~4.19).
So $V \approx 4.19 \times 10^{24} \times 10^3$ ? No — careful: $r = 7.0 \times 10^8$ , so $r^3 = 343 \times 10^{24} = 3.43 \times 10^{26}$ .

Then $V = \frac{4}{3}\pi \times 3.43 \times 10^{26} \approx 1.44 \times 10^{27}\ \text{m}^3$ .
Then density:

\rho = \frac{2.0 \times 10^{30}}{1.44 \times 10^{27}} \approx 1.39 \times 10^3\ \text{kg/m}^3

5. ✗ Not comparing with known density ranges

Mistake: Computing $\rho \approx 1400\ \text{kg/m}^3$ but not knowing what it means.

How to avoid: Memorise reference values:

Gases: $\sim 1\ \text{kg/m}^3$ (air at STP)
Liquids: $\sim 1000\ \text{kg/m}^3$ (water)
Solids: $\sim 2000\!-\!8000\ \text{kg/m}^3$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2024Set D-21 markMCQ
$^{27}$
›Reveal solutionSolution
The ratio of volume to surface area for any spherical nucleus is $R/3$ , where $R = R_0 A^{1/3}$ . For Al $^{27}$ , $A=27$ , so $R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15}$ m, giving a ratio of $1.2 \times 10^{-15}$ m. The correct option is (D).

The key idea here is that a nucleus is modeled as a uniform sphere. For any sphere, the ratio of volume to surface area is simply $R/3$ — a clean geometric fact. Once you know the nuclear radius formula, the rest is arithmetic.

The nuclear radius is given by $R = R_0 A^{1/3}$ , where $R_0 = 1.2 \times 10^{-15}$ m and $A$ is the mass number. For Al $^{27}$ , $A = 27$ , so $A^{1/3} = 3$ . That gives $R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15}$ m.

Now, for a sphere:

Volume $V = \frac{4}{3}\pi R^3$

Surface area $S = 4\pi R^2$

The ratio $\frac{V}{S} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3}$ .

So the ratio is simply $\frac{3.6 \times 10^{-15}}{3} = 1.2 \times 10^{-15}$ m. …
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KCET 2019Set A-11 markMCQ
$\alpha$
›Reveal solutionSolution
In a head-on collision, the $\alpha$ -particle is aimed directly at the nucleus, so the perpendicular distance between the initial velocity line and the nucleus — the impact parameter — is exactly zero.

The key idea here is the definition of impact parameter in Rutherford's scattering experiment. The impact parameter $b$ is the perpendicular distance between the initial velocity vector of the $\alpha$ -particle and the centre of the target nucleus. It tells you how "off-centre" the collision is.

For a head-on collision, the $\alpha$ -particle is aimed straight at the nucleus. That means the line of the initial velocity passes directly through the centre of the nucleus. The perpendicular distance from that line to the centre is therefore zero.

Recall the definition: Impact parameter $b$ = distance of closest approach if the nucleus were not there — it's the miss distance. Mathematically, if the initial velocity is along a line, $b$ is the perpendicular distance from the nucleus centre to that line.

Head-on means zero miss distance: When the $\alpha$ -particle is fired straight at the nucleus, there is no "miss" — it's coming right at it. So $b = 0$ .

Consequences: For $b = 0$ , the $\alpha$ -particle experiences the maximum repulsive force (Coulomb force) and comes to rest momentarily at the distance of closest approach before being repelled back. This is the case that gives the largest scattering angle ( $180^\circ$ ). …
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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

Why "hot plasma" doesn't automatically mean "low density"

Computing the average density

Interpreting the result …

Method: Density Estimation from Mass and Volume

Steps

Common Mistakes in Atomic Volume / Density Calculation (Sun Problem)

1. ✗ Using wrong formula for volume of a sphere

2. ✗ Forgetting to cube the radius correctly

3. ✗ Mixing up units (km vs m)

4. ✗ Arithmetic errors in powers of 10

5. ✗ Not comparing with known density ranges

Why "hot plasma" doesn't automatically mean "low density"

Computing the average density

Interpreting the result …

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