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Physics / Chapter 1: Measurement / Section 1-4

Length — Question 1

Section 1-4 · Physics pages 72–73

Question 1 Atomic lengths in a scale model For an illustrative scale model, assign an atom a diameter da = 0.10 nm and its nucleus a diameter dn = 5.0 fm. These are stipulated representative sizes, not a claim that all atoms and nuclei have these diameters. Use exact 1 nm = 10−9 m and 1 fm = 10−15 m. Nucleus Atom da = 0.10 nm dn = 5.0 fm Separate magnifications: the nucleus is enlarged and the drawing is not to scale. (a) Express both diameters in metres and find da/dn. A student compares the printed numbers 0.10 and 5.0 and concludes that the nucleus is larger. Explain the failure in that comparison. (b) Enlarge both lengths by the same factor until the model atom has diameter 100 m. Find the model nucleus diameter and choose a convenient unit to communicate its size. (c) If both regions are represented by spheres, use V = πd3/6 to find Vn/Va. Why is the volume fraction not the same as the diameter fraction? (d) Does this geometric model establish the material density of the nucleus, or prove that an atom behaves like a solid ball? Explain what additional information or physics would be needed. Use the stipulated sizes as exact within the model, and distinguish conclusions about scale from conclusions about physical structure.
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Question 1 - Solution Atomic lengths in a scale model Nucleus Atom da = 0.10 nm dn = 5.0 fm (a) Compare lengths in the same unit. da = 1.0 × 10−10 m, dn = 5.0 × 10−15 m, da dn = 2.0 × 104 = 20 000 . The atom’s assigned diameter is twenty thousand times the nucleus’s. The numerical values 0.10 and 5.0 multiply units differing by a factor of a million; their order alone says nothing about the physical sizes. (b) Preserve the linear ratio in the enlarged model. The enlargement factor is (100 m)/(10−10 m) = 1012. Thus the model nucleus has diameter (5.0 × 10−15 m)(1012) = 0.0050 m = 5.0 mm . A millimetre-scale nucleus inside a hundred-metre-scale atom communicates the large difference in length scales. (c) Volume carries three powers of length. For the two stipulated spheres the common factor π/6 cancels: Vn Va = (dn da )3 = ( 1 20 000 )3 = 1.25 × 10−13 . The diameter fraction is 5.0×10−5; the volume fraction cubes it because all three spatial dimensions scale together. (d) State the model’s limits. ρn = mn Vn = 6mn πd3n ; dn alone does not determine mn. No mass was supplied, so no mass density follows from these lengths alone. Nor does using spheres as geometric reference regions imply that the atom is a uniform solid sphere. Predictions about matter distribution or mechanical behavior require an appropriate physical model and further information. A drawing or enlargement preserves stipulated lengths; it does not determine the underlying structure.

Question 1

Atomic lengths in a scale model

For an illustrative scale model, assign an atom a diameter da=0.10nmd_a=0.10\ \mathrm{nm} and its nucleus a diameter dn=5.0fmd_n=5.0\ \mathrm{fm}. These are stipulated representative sizes, not a claim that all atoms and nuclei have these diameters. Use exact 1nm=10−9m1\ \mathrm{nm}=10^{-9}\ \mathrm{m} and 1fm=10−15m1\ \mathrm{fm}=10^{-15}\ \mathrm{m}.

Diagram shown in the original SVG worksheet.

  1. Express both diameters in metres and find da/dnd_a/d_n. A student compares the printed numbers 0.10 and 5.0 and concludes that the nucleus is larger. Explain the failure in that comparison.

  2. Enlarge both lengths by the same factor until the model atom has diameter 100m100\ \mathrm{m}. Find the model nucleus diameter and choose a convenient unit to communicate its size.

  3. If both regions are represented by spheres, use V=πd3/6V=\pi d^3/6 to find Vn/VaV_n/V_a. Why is the volume fraction not the same as the diameter fraction?

  4. Does this geometric model establish the material density of the nucleus, or prove that an atom behaves like a solid ball? Explain what additional information or physics would be needed.

Use the stipulated sizes as exact within the model, and distinguish conclusions about scale from conclusions about physical structure.

Question 1 - Solution

Atomic lengths in a scale model

Diagram shown in the original SVG worksheet.

(a) Compare lengths in the same unit.

da=1.0×10−10m,dn=5.0×10−15m,dadn=2.0×104=20000.d_a=1.0\times10^{-10}\ \mathrm{m},\qquad d_n=5.0\times10^{-15}\ \mathrm{m},\qquad \boxed{\frac{d_a}{d_n}=2.0\times10^4=20\,000}. The atom’s assigned diameter is twenty thousand times the nucleus’s. The numerical values 0.10 and 5.0 multiply units differing by a factor of a million; their order alone says nothing about the physical sizes.

(b) Preserve the linear ratio in the enlarged model.

The enlargement factor is (100m)/(10−10m)=1012(100\ \mathrm{m})/(10^{-10}\ \mathrm{m})=10^{12}. Thus the model nucleus has diameter (5.0×10−15m)(1012)=0.0050m=5.0mm.(5.0\times10^{-15}\ \mathrm{m})(10^{12}) =0.0050\ \mathrm{m}=\boxed{5.0\ \mathrm{mm}}. A millimetre-scale nucleus inside a hundred-metre-scale atom communicates the large difference in length scales.

(c) Volume carries three powers of length.

For the two stipulated spheres the common factor π/6\pi/6 cancels: VnVa=(dnda)3=(120000)3=1.25×10−13.\frac{V_n}{V_a}=\left(\frac{d_n}{d_a}\right)^3 =\left(\frac{1}{20\,000}\right)^3 =\boxed{1.25\times10^{-13}}. The diameter fraction is 5.0×10−55.0\times10^{-5}; the volume fraction cubes it because all three spatial dimensions scale together.

(d) State the model’s limits.

ρn=mnVn=6mnπdn3;dn alone does not determine mn.\rho_n=\frac{m_n}{V_n}=\frac{6m_n}{\pi d_n^3};\qquad d_n\text{ alone does not determine }m_n.

No mass was supplied, so no mass density follows from these lengths alone. Nor does using spheres as geometric reference regions imply that the atom is a uniform solid sphere. Predictions about matter distribution or mechanical behavior require an appropriate physical model and further information. A drawing or enlargement preserves stipulated lengths; it does not determine the underlying structure.

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