Q.In the Rutherford's nuclear model of the atom, the nucleus (radius about 10−15 m) is analogous to the sun about which the electron moves in orbit (radius ≈10−10 m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is? The radius of earth's orbit is about 1.5×1011 m. The radius of sun is taken as 7×108 m.
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 1×10−10 m (1 angstrom)
- Radius of a typical nucleus: about 1×10−15 m (1 femtometre)
The ratio is
rnucleusratom≈10−15 m10−10 m=105
The atom is about one hundred thousand times wider than its nucleus.
Making the number imaginable
A factor of 105 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 cm across), then the atom would be a sphere roughly 105 times bigger — around 1 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 105 in radius packs it into a volume 1015 times smaller. That is why nuclear matter has an almost unimaginable density — on the order of 1017 kg/m3 — while the atom as a whole is light and airy.
Only the linear sizes scale by 105. Areas scale as the square (1010) and volumes as the cube (1015). Keep track of which quantity you are comparing before you quote a ratio.
Why this matters for the exam
- The huge atom-to-nucleus size ratio (∼105) is the direct evidence that atoms are mostly empty and that positive charge is concentrated in a tiny core.
- It explains Rutherford's key observation: most alphas undeflected, a rare few scattered through large angles.
- Remember the two benchmark sizes — atom ∼10−10 m, nucleus ∼10−15 m — and the pea-in-a-stadium picture that follows from them.
Do not confuse linear scale with volume scale. Saying "the nucleus is 105 times smaller" refers to radius; by volume it is smaller by a factor of about 1015.
The pea-in-a-stadium scale analogy is a well-known way NCERT and CBSE Class 12 Physics textbooks help students visualise the atom-to-nucleus size ratio from the Atoms chapter, and it shows up often in "atomic size vs nuclear size comparison" and "Rutherford's gold foil experiment important questions" searches. This intuition-building concept is a favourite in board-exam short-answer questions precisely because it tests understanding rather than pure calculation.
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 m across while the whole atom is about 10−10 m — the atom is roughly 100,000 times wider than its nucleus, so it is almost entirely empty space.
The Sizes Involved
- Atomic radius: ratom≈10−10 m (1 angstrom).
- Nuclear radius: rnucleus≈10−15 m (1 femtometre).
The ratio of diameters is:
rnucleusratom≈10−1510−10=105
Bringing It to Human Scale
Imagine blowing the nucleus up to the size of a cricket ball (radius ≈3.5 cm). To keep the same ratio, the electrons would orbit at:
0.035 m×105=3500 m≈3.5 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.
Why It Matters
The volume ratio scales as the cube of the length ratio, (105)3=1015, so the nucleus occupies only about one part in 1015 of the atom's volume yet holds over 99.9% of its mass. This is exactly why most of Rutherford's alpha particles passed straight through the gold foil, while a rare few — those aimed almost dead-on at a nucleus — bounced sharply back.
The analogy captures relative sizes only; electrons are not little balls on tracks but a quantum probability cloud.
Compare the two "orbit-radius to central-body-radius" ratios. For the atom, rorbit/rnucleus=10−10/10−15=105. For the real solar system, Rorbit/Rsun=1.5×1011/7×108≈214. To make the solar system atom-like (same 105 ratio) with the Sun's radius fixed, the orbit would need R′=105×7×108=7×1013 m, about 470 times the real 1.5×1011 m.
The Earth would be much farther from the Sun — its orbit would have to be about 7×1013 m, roughly 470 times its actual radius, because the nucleus is far smaller (relative to its orbit) than the Sun is relative to Earth's orbit.
The atom is far emptier than the solar system, so matching its proportions pushes Earth's orbit out to about 7×1013 m — roughly 470 times its real radius. The Earth would be much farther from the Sun.
Concept understanding
The scale of an orbiting system is captured by the ratio of the orbit radius to the radius of the central body. We compare that ratio for the atom with the same ratio for the solar system.
Working it out
1. The atom's ratio.
rnucleusrorbit=10−1510−10=105.
The electron orbits at 105 times the nuclear radius.
2. The real solar system's ratio.
RsunRorbit=7×1081.5×1011≈2.14×102≈214.
3. Rescale to be atom-like. Keeping the Sun (the nucleus analogue) fixed, the orbit must satisfy Rorbit′/Rsun=105:
Rorbit′=105×7×108=7×1013 m.
4. Compare with the true orbit.
RorbitRorbit′=1.5×10117×1013≈4.7×102≈470.
Because the nucleus is tinier relative to the electron's orbit (105) than the Sun is relative to Earth's orbit (214), reproducing the atomic proportion forces the orbit to swell by a factor of about 470.
The Earth would be much farther from the Sun. To match the atom's proportions its orbit would have to grow to about 7×1013 m — roughly 470 times the actual 1.5×1011 m.
Method: Direct Proportion (Scale Factor Comparison)
The core idea is that the atom and the solar system are being treated as scale models of each other. In the atom, the nucleus is the “sun” and the electron’s orbit is the “earth’s orbit.” We find the ratio of orbit radius to nucleus radius in the atom, then apply that same ratio to the sun to see what the earth’s orbit would be if the solar system were scaled the same way.
Steps
- Find the scale factor in the atom. The electron orbits at radius 10−10 m around a nucleus of radius 10−15 m. The ratio is:
nucleus radiusorbit radius=10−1510−10=105
So the orbit is 105 times larger than the nucleus.
- Apply the same scale factor to the solar system. The sun’s radius is given as 7×108 m. If the earth’s orbit were proportioned like the atom, its radius would be:
scaled orbit radius=(sun’s radius)×105=(7×108)×105=7×1013 m
- Compare with the actual earth–sun distance. The actual radius of earth’s orbit is 1.5×1011 m. Since 7×1013 m is much larger than 1.5×1011 m, the scaled orbit is farther away.
The atom’s electron orbit is 105 times the nucleus size. Applying that to the sun gives an orbit 7×1013 m, which is about 467 times the actual earth–sun distance.
Final answer:
The earth would be farther away from the sun than it actually is.
Common Mistakes & How to Avoid Them
Mistake 1: Confusing which ratio to compare
Students often compare the nucleus radius to the electron orbit radius directly, or compare the sun's radius to the earth's orbit radius — without realising the analogy works by matching the same kind of ratio on both sides.
How to avoid: The analogy says: nucleus : electron orbit is like sun : earth's orbit. So you must set up:
electron orbit radiusnucleus radius=earth’s orbit radius (new)sun radius
You are not comparing nucleus to sun directly. You are comparing the proportions of the two systems.
Write the two ratios side by side before plugging numbers.
Atom: 10−1010−15=10−5
Solar system: Rnew7×108 — set them equal.
Mistake 2: Forgetting to solve for the new earth-sun distance
Many students compute the ratio 10−5 and then stop, or they multiply the wrong quantities. They might calculate 7×108×10−5 instead of dividing.
How to avoid: Once you set up the proportion:
10−1010−15=Rnew7×108
Cross-multiply carefully:
10−15×Rnew=10−10×7×108
Rnew=10−157×10−2=7×1013 m
A common slip: writing 10−15×R=7×108×10−10 is correct, but then dividing 7×108 by 10−5 instead of 7×10−2 by 10−15. Track your exponents step by step.
Mistake 3: Misinterpreting "closer or farther"
After finding Rnew=7×1013 m, students compare it to the sun's radius instead of the actual earth-sun distance (1.5×1011 m).
How to avoid: The question asks: compared to the actual earth-sun distance, is the new distance larger or smaller?
Actual: 1.5×1011 m
New: 7×1013 m
Since 7×1013>1.5×1011, the earth would be farther away.
Always re-read the question's last sentence. It tells you exactly which two numbers to compare at the end.
Mistake 4: Using the wrong units or forgetting powers of ten
Students sometimes treat 10−15 and 10−10 as if they were 10−15 m and 10−10 m but then drop the exponents when dividing, or misplace decimal points with 7×108.
How to avoid: Write every number in scientific notation before calculating. Do not approximate 10−15/10−10 as 10−5 in your head without writing it down — one slip and the answer is off by a factor of 10.
Mistake 5: Thinking the analogy means "same size" not "same proportion"
Some students assume the nucleus is the sun and the electron is the earth, so they directly compare 10−15 to 7×108 and conclude the atom is much smaller — missing the point entirely.
How to avoid: The word "analogous" means the relationship is similar, not the sizes. The atom's nucleus is tiny compared to its orbit; the question asks: if the solar system had that same ratio, what would happen?
| System | Central object radius | Orbital radius | Ratio (centre/orbit) |
|--------|----------------------|----------------|----------------------|
| Atom | 10−15 m | 10−10 m | 10−5 |
| Solar system (actual) | 7×108 m | 1.5×1011 m | ≈4.7×10−3 |
| Solar system (scaled) | 7×108 m | 7×1013 m | 10−5 |
The scaled solar system has a much larger orbit because the sun stays the same size but the ratio must shrink to match the atom's extreme proportion.
Final answer: The earth would be farther away from the sun than it actually is — at a distance of 7×1013 m instead of 1.5×1011 m.
- KCET 2021Set B-21 markMCQQ.In a nuclear reactor heavy nuclei is not used as moderators because (A) They will break up (B) Elastic collision of neutrons with heavy nuclei will not slow them down. (C) The net weight of the reactor would be unbearably high (D) Substances with heavy nuclei do not occur in liquid or gaseous state at room temperature.
›Reveal solutionSolution
A moderator must slow down neutrons via elastic collisions, which requires the moderator nuclei to have a mass comparable to the neutron. Heavy nuclei are too massive to absorb enough kinetic energy in a single collision, making them ineffective — so option (B) is correct.
The key idea here is how a moderator works. In a nuclear reactor, the fission of uranium-235 produces fast neutrons (with energies around 1–2 MeV). These fast neutrons are not very efficient at causing further fission in uranium-235 — they are more likely to be captured without fission, or to escape. To sustain a chain reaction, we need to slow these neutrons down to thermal energies (about 0.025 eV), where the fission cross-section is much larger. That’s the moderator’s job.
A moderator slows neutrons by elastic collisions. Think of it like billiard balls: when a moving ball hits a stationary one, the lighter the stationary ball, the more speed it takes away from the moving ball. The best energy transfer happens when the two masses are equal. A hydrogen nucleus (a proton) has almost the same mass as a neutron, so a neutron can lose up to 100% of its energy in a single head-on collision. A carbon nucleus (mass 12 u) is heavier, but still light enough to slow neutrons reasonably well. But a heavy nucleus — say, lead (mass 207 u) — is like a bowling ball hitting a wall: the wall barely moves, so the ball keeps almost all its speed.
Let’s go through the options one by one.
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Option (A): "They will break up"
Heavy nuclei are stable against breakup from neutron collisions at these energies. A neutron colliding with a heavy nucleus does not have enough energy to cause nuclear fission or disintegration — that requires much higher energies or specific isotopes. So this is not the reason.
-
Option (B): "Elastic collision of neutrons with heavy nuclei will not slow them down"
This is the correct physics. In an elastic collision, the fraction of kinetic energy transferred from a neutron (mass m) to a stationary nucleus (mass M) is given by:
EΔE=(m+M)24mMcos2θ
where θ is the scattering angle in the centre-of-mass frame. The maximum transfer (when θ=0) is:
EΔEmax=(m+M)24mM
For a heavy nucleus, M≫m, so this fraction becomes very small:
(m+M)24mM≈M4m≪1
For example, with a lead nucleus (M≈207 u), the maximum energy transfer is only about 4/207≈1.9% per collision. It would take hundreds of collisions to slow a neutron down to thermal energies — impractical. In contrast, with hydrogen (M=1 u), the maximum transfer is 100%, and with carbon (M=12 u), it’s about 28%. Heavy nuclei simply cannot slow neutrons efficiently.
TipA quick way to remember: the best moderator has nuclei with mass close to the neutron’s. That’s why light water (hydrogen), heavy water (deuterium), and graphite (carbon) are common moderators — all have low mass numbers.
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Option (C): "The net weight of the reactor would be unbearably high"
While heavy nuclei are dense, weight is not the primary concern — reactors are already massive structures. More importantly, even if weight were acceptable, the moderator would still fail to slow neutrons. So this is not the fundamental reason.
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Option (D): "Substances with heavy nuclei do not occur in liquid or gaseous state at room temperature"
This is factually incorrect. Many heavy elements exist as liquids (mercury, lead at high temperature) or gases (radon, xenon). But even if they did, the state of matter is not the issue — the physics of elastic collisions is.
Watch outA common mistake is to think heavy nuclei absorb neutrons or break apart. They don’t — the problem is purely about energy transfer in elastic collisions, not absorption or instability.
✓Final answerThe correct option is (B) — elastic collisions of neutrons with heavy nuclei will not slow them down effectively.
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