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Q.If r1r_1 and r2r_2 are the radii of atomic nuclei of mass numbers 6464 and 2727 respectively, then the value of r1r2\dfrac{r_1}{r_2} is : (A) 11 (B) 43\dfrac{4}{3} (C) 34\dfrac{3}{4} (D) 2764\dfrac{27}{64}

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Nuclear radius scales as the cube root of mass number: r∝A1/3r \propto A^{1/3}. Taking the ratio gives r1r2=(6427)1/3=43\frac{r_1}{r_2} = \left(\frac{64}{27}\right)^{1/3} = \frac{4}{3}.

The size of an atomic nucleus is not arbitrary. Experiments—particularly Rutherford scattering and later electron-scattering studies—revealed that nuclear radius follows a beautiful empirical law that connects it directly to how many nucleons (protons and neutrons) are packed inside.

The key insight is that nuclear matter has roughly constant density. Think of nucleons as incompressible spheres packed together: if you double the number of nucleons, you double the volume, not the radius. Since volume scales as r3r^3, the radius must scale as the cube root of the number of nucleons.

r=r0A1/3r = r_0 A^{1/3}

where AA is the mass number (total nucleons) and r0≈1.2 fmr_0 \approx 1.2 \text{ fm} is a constant.

Now let's find the ratio of the two nuclear radii.

  1. Write the radius for each nucleus For the nucleus with mass number A1=64A_1 = 64:

r1=r0(64)1/3r_1 = r_0 (64)^{1/3}

For the nucleus with mass number A2=27A_2 = 27:

r2=r0(27)1/3r_2 = r_0 (27)^{1/3}

  1. Form the ratio When we divide r1r_1 by r2r_2, the constant r0r_0 cancels: …

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