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Q13.4:
Obtain approximately the ratio of the nuclear radii of the gold isotope \(^{197}_{79}Au\) and the silver isotope \(^{107}_{47}Ag\).

Solution :

Given Variables & Initial Setup

To determine the ratio of the nuclear radii, we identify the mass numbers ($A$) of the respective isotopes provided:

  • Gold isotope ($^{197}_{79}\text{Au}$): Mass number $A_{Au} = 197$
  • Silver isotope ($^{107}_{47}\text{Ag}$): Mass number $A_{Ag} = 107$

Theoretical Foundation: Empirical Relation for Nuclear Radius

The radius of a nucleus ($R$) is empirically related to its mass number ($A$) by the following formula: $$R = R_0 A^{1/3}$$ [Per the empirical formula for nuclear density distribution, where $R_0 \approx 1.2 \times 10^{-15} \text{ m}$ (or $1.2 \text{ fm}$), and the nuclear density is approximately constant across all nuclei].

Step 1: Formulating the Ratio Equation

We define the ratio of the radius of Gold ($R_{Au}$) to the radius of Silver ($R_{Ag}$): $$\frac{R_{Au}}{R_{Ag}} = \frac{R_0 (A_{Au})^{1/3}}{R_0 (A_{Ag})^{1/3}}$$ [By algebraic substitution into the nuclear radius law].

Step 2: Simplifying the Expression

Canceling the constant term $R_0$ from the numerator and denominator: $$\frac{R_{Au}}{R_{Ag}} = \left( \frac{A_{Au}}{A_{Ag}} \right)^{1/3}$$ Substituting the given mass numbers: $$\frac{R_{Au}}{R_{Ag}} = \left( \frac{197}{107} \right)^{1/3}$$

Step 3: Numerical Computation

Performing the division inside the parentheses: $$\frac{197}{107} \approx 1.8411$$ Now, calculating the cube root: $$\left( 1.8411 \right)^{1/3} \approx 1.225$$ [By numerical evaluation of fractional exponents].

Visual Representation of Nuclear Scale

Gold (A=197) Silver (A=107) R_Au R_Ag

Final Solution: The ratio of the nuclear radius of Gold to Silver is approximately 1.22 : 1.


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