Which of the following properties of atom could be explained correctly by Thomson Model of atom? (i) Overall n

NCERT Exemplar · Q6

Q. Which of the following properties of atom could be explained correctly by Thomson Model of atom?

(i) Overall neutrality of atom.

(ii) Spectra of hydrogen atom.

(iii) Position of electrons, protons and neutrons in atom.

(iv) Stability of atom.

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Concept understanding — Scale Analogy

Scale Analogy

Why we need an analogy at all

Rutherford's gold-foil experiment revealed something almost impossible to picture: the atom's entire positive charge and nearly all of its mass sit inside a nucleus that is fantastically smaller than the atom around it. The numbers are so extreme that our everyday intuition breaks down — so physicists reach for a scale analogy: blow the atom up to a size we can imagine, and see where the nucleus ends up.

The intuition: the atom is mostly empty space

Most alpha particles fired at the gold foil passed straight through, barely deflected. Only about 1 in 8000 bounced back sharply. The only way to explain this is that the atom is overwhelmingly empty, with a tiny, dense, positively charged core that the occasional alpha particle scores a near-direct hit on.

So how tiny is "tiny"? Compare the two sizes:

Radius of a typical atom: about $1 \times 10^{-10}\ \text{m}$ ( $1\ \text{angstrom}$ )
Radius of a typical nucleus: about $1 \times 10^{-15}\ \text{m}$ ( $1\ \text{femtometre}$ )

The ratio is

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

The atom is about one hundred thousand times wider than its nucleus.

Making the number imaginable

A factor of $10^5$ is just a symbol on paper. The scale analogy converts it into something the mind can hold:

If the nucleus were the size of a pea (about $1\ \text{cm}$ across), then the atom would be a sphere roughly $10^5$ times bigger — around $1\ \text{km}$ across. The pea would sit alone at the centre of a stadium-sized region of empty space, with the electrons whirling somewhere out near the edge.
Equivalently, if the atom were scaled up to the size of a large sports ground, the nucleus would be no bigger than a grain of sand at the centre-spot.

Either picture drives home the same point: an atom is almost entirely empty space, which is exactly why nearly every alpha particle sailed through the foil undeflected.

Density: the flip side of the analogy

The scale analogy also warns us about density. Nearly the whole mass of the atom is squeezed into that pin-point nucleus. Because volume grows as the cube of the radius, shrinking the mass-holder by $10^5$ in radius packs it into a volume $10^{15}$ times smaller. That is why nuclear matter has an almost unimaginable density — on the order of $10^{17}\ \text{kg/m}^3$ — while the atom as a whole is light and airy. …

Why this formula?

Scale Analogy

One of the hardest facts to picture in atomic physics is just how empty an atom is. Rutherford's scattering experiment showed that almost all the mass sits in a tiny central nucleus, with the electrons far outside. A scale analogy makes the numbers vivid.

The nucleus is about $10^{-15}\ \text{m}$ across while the whole atom is about $10^{-10}\ \text{m}$ — the atom is roughly 100,000 times wider than its nucleus, so it is almost entirely empty space.

The Sizes Involved

Atomic radius: $r_{\text{atom}} \approx 10^{-10}\ \text{m}$ (1 angstrom).
Nuclear radius: $r_{\text{nucleus}} \approx 10^{-15}\ \text{m}$ (1 femtometre).

The ratio of diameters is:

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

Bringing It to Human Scale

Imagine blowing the nucleus up to the size of a cricket ball (radius $\approx 3.5\ \text{cm}$ ). To keep the same ratio, the electrons would orbit at:

0.035\ \text{m}\times 10^{5} = 3500\ \text{m} \approx 3.5\ \text{km}

So a nucleus the size of a ball at the centre of a stadium would have its electrons drifting kilometres away — and the space in between is vacuum. …

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Thomson’s model (the “plum pudding” model) could explain the overall neutrality of an atom and its stability in a static sense, but it completely failed to explain atomic spectra or the actual arrangement of subatomic particles. The correct options are (A) and (D).

The Concept First — The Scale Analogy

Imagine a spherical, positively charged jelly, with tiny negative raisins (electrons) embedded inside it. That’s Thomson’s atom. The key idea is that the positive charge is spread uniformly throughout the whole sphere, and the electrons are sprinkled in just the right way to make the atom electrically neutral. This model was the first to propose that atoms are not indivisible — they contain smaller charged particles. But it was a static picture, like a snapshot. It couldn’t explain why atoms emit light of specific colours (spectra) or where the heavy nucleus sits. Let’s check each property.

Step-by-Step Reasoning

1. Overall neutrality of atom.

Thomson’s model explicitly states that the total negative charge of the electrons exactly balances the total positive charge of the sphere. So the atom as a whole has zero net charge. This was a direct and correct prediction of the model.

Tip

This is the simplest check: if the number of electrons equals the number of positive charge units, the atom is neutral. Thomson’s model was built on that idea.

2. Spectra of hydrogen atom.

Thomson’s model gives no mechanism for electrons to occupy specific energy levels. Electrons are just stuck in the positive jelly like raisins in a pudding. There’s no quantization of energy, no orbits, no way to explain why hydrogen emits only certain wavelengths of light. The model predicts a continuous range of possible energies, which contradicts the observed line spectra.

Watch out

A common mistake is to think Thomson’s model could explain spectra because it had electrons. But having electrons is not enough — you need discrete energy levels, which only came with Bohr’s model.

3. Position of electrons, protons and neutrons in atom.

Thomson’s model has no protons or neutrons — it only has a diffuse positive sphere and electrons. The concept of a tiny, dense nucleus (protons and neutrons) was discovered later by Rutherford. So Thomson’s model cannot account for the actual positions of these particles at all. …

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Q.Which of the following properties of atom could be explained correctly by Thomson Model of atom?

The Concept First — The Scale Analogy

Step-by-Step Reasoning

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