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Exercises · 13.11

Q.Obtain approximately the ratio of the nuclear radii of the gold isotope 79197Au^{197}_{79}\text{Au} and the silver isotope 47107Ag^{107}_{47}\text{Ag}.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★
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Nuclear radius scales as R∝A1/3R \propto A^{1/3}, so the ratio of radii for gold (A=197A=197) and silver (A=107A=107) is 197/1073≈1.23\sqrt[3]{197/107} \approx 1.23.

The key idea is that nuclear density is roughly constant for all nuclei — a fact known from scattering experiments. Since mass number AA is proportional to volume, and volume scales as R3R^3, we get R∝A1/3R \propto A^{1/3}. This is one of the most reliable scaling laws in nuclear physics.

Let’s work through it.

  1. The nuclear radius formula For any nucleus, the radius is given by

R=R0A1/3R = R_0 A^{1/3}

where R0≈1.2×10−15 mR_0 \approx 1.2 \times 10^{-15}\,\text{m} (a constant). The exact value of R0R_0 cancels out when we take a ratio, so we don’t need it here.

  1. Write the ratio For gold: RAu=R0(197)1/3R_{\text{Au}} = R_0 (197)^{1/3} For silver: RAg=R0(107)1/3R_{\text{Ag}} = R_0 (107)^{1/3} Dividing:

RAuRAg=(197)1/3(107)1/3=(197107)1/3\frac{R_{\text{Au}}}{R_{\text{Ag}}} = \frac{(197)^{1/3}}{(107)^{1/3}} = \left(\frac{197}{107}\right)^{1/3}

  1. Approximate the fraction 197/107≈1.841197/107 \approx 1.841. Now take the cube root. A quick mental check: 1.23=1.7281.2^3 = 1.728, 1.253=1.9531.25^3 = 1.953. So the cube root lies between 1.2 and 1.25. More precisely, 1.233=1.23×1.23×1.23=1.5129×1.23≈1.8611.23^3 = 1.23 \times 1.23 \times 1.23 = 1.5129 \times 1.23 \approx 1.861 — very close to 1.841. So (197107)1/3≈1.23\left(\frac{197}{107}\right)^{1/3} \approx 1.23 …

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