Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{79}^{197}{Au}$ and the silver isotope ${}_{47}^{107}{Ag} .$

Solution

Using the relation between the radius of nucleus and atomic mass,
$R \approx A^{1 / 3}$

Atomic mass of gold, A₁ = 197
Atomic mass of silver, A₂ = 107

$∴$ $\frac{R_{1}}{R_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$

$= {(\frac{197}{107})}^{1 / 3} = (1.84)^{1 / 3}$

Now, taking log on both sides,

$\Rightarrow$ $\log_{10} (\frac{R_{1}}{R_{2}}) = \log_{10} (1.84)^{1 / 3}$

$\Rightarrow$ $\log_{10} (\frac{R_{1}}{R_{2}}) = \frac{1}{3} \log_{10} (1.84)$

$= \frac{1}{3} \times 0.2648 = 0.08827$

$\Rightarrow$ $\frac{R_{1}}{R_{2}} = antilog (0.08827)$
= 1.23, which is the required ratio of the nucleii.

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Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{79}^{197}{Au}$ and the silver isotope ${}_{47}^{107}{Ag} .$

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Wave Optics

Obtain approximately the ratio of the nuclear radii of the gold isotope Au79197 and the silver isotope Ag47107.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{79}^{197}{Au}$ and the silver isotope ${}_{47}^{107}{Ag} .$