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.
- CBSE 2020Set ANNUAL5 marksQ.(a) Explain Rutherford model of an atom. [3M].(b) Calculate the mass of a photon with wavelength 2.5 A°. [2M]
›Reveal solutionSolution
[!TLDR]
(a) Based on his alpha-particle scattering experiment (thin gold foil bombarded with alpha particles), Rutherford observed that most alpha particles passed straight through undeflected, a few were deflected at small angles, and a very small fraction (about 1 in 20,000) bounced back at angles close to 180°. From this he concluded: (i) most of the atom's volume is empty space; (ii) almost all the mass and the entire positive charge of the atom is concentrated in a very small, dense central region called the nucleus; (iii) electrons revolve around the nucleus in circular orbits at relatively large distances; (iv) the size of the nucleus is extremely small compared to the size of the atom. (This model could not explain the stability of the atom or its line spectrum.) (b) m = h/(λc) = (6.626×10^-34 J s) / (2.5×10^-10 m × 3×10^8 m/s) = 6.626×10^-34 / 7.5×10^-2 ≈ 8.83×10^-33 kg.
Method
(a) This is the standard account of Rutherford's alpha-scattering experiment and its atomic-model conclusions. (b) For a photon, E = hc/λ and E = mc², so m = h/(λc); substituting h, c and λ=2.5Å=2.5×10^-10 m gives the photon's (relativistic) mass.
[!ANSWER]
(a) Based on his alpha-particle scattering experiment (thin gold foil bombarded with alpha particles), Rutherford observed that most alpha particles passed straight through undeflected, a few were deflected at small angles, and a very small fraction (about 1 in 20,000) bounced back at angles close to 180°. From this he concluded: (i) most of the atom's volume is empty space; (ii) almost all the mass and the entire positive charge of the atom is concentrated in a very small, dense central region called the nucleus; (iii) electrons revolve around the nucleus in circular orbits at relatively large distances; (iv) the size of the nucleus is extremely small compared to the size of the atom. (This model could not explain the stability of the atom or its line spectrum.) (b) m = h/(λc) = (6.626×10^-34 J s) / (2.5×10^-10 m × 3×10^8 m/s) = 6.626×10^-34 / 7.5×10^-2 ≈ 8.83×10^-33 kg.
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