Question

The graph of $\ln \left( \dfrac{R}{{{R}_{0}}} \right)$ versus $\ln (A)$of nucleus is:
(${{R}_{0}}$ is a constant $1.2\times {{10}^{-12}}$m, R is radius and A is the atomic number)
(A) a straight line
(B) a parabola
(C) an eclipse
(D) a circle

Answer 1

Hint: We will use the relation between the radius and atomic number of an atom. This relation is given as, the radius of any atom is directly proportional to its atomic number to the power of one by three. In this equation, the proportionality constant is given the term of ${{R}_{0}}$. We shall use this relation to proceed ahead in our solution.

Complete step-by-step answer:
Let the radius of any atom having an atomic number A be denoted by R. Then, they are related as follows:
$\Rightarrow R\propto {{\left( A \right)}^{\dfrac{1}{3}}}$
Which is further written as :
$\Rightarrow R={{R}_{0}}{{\left( A \right)}^{\dfrac{1}{3}}}$ [Let this expression be equation number (1)]
Where, ${{R}_{0}}$is the constant of proportionality and its value is $1.2\times {{10}^{-12}}$m.
Taking logarithm to the base ‘e’ on both sides of equation number (1), we get:
$\Rightarrow \ln R=\ln [{{R}_{0}}{{\left( A \right)}^{\dfrac{1}{3}}}]$
Using the property of logarithm, that is:
$\Rightarrow \ln (AB)=\ln A+\ln B$
We get:
$\begin{align}
  & \Rightarrow \ln R=\ln {{R}_{0}}+\dfrac{1}{3}\ln A \\
 & \Rightarrow \ln R-\ln {{R}_{0}}=\dfrac{1}{3}\ln A \\
\end{align}$
Now using the property of logarithm:
$\Rightarrow \ln A-\ln B=\ln \left( \dfrac{A}{B} \right)$
We get the above equation as:
$\Rightarrow \ln \left( \dfrac{R}{{{R}_{0}}} \right)=\dfrac{1}{3}\ln A$
Assuming the term $\ln \left( \dfrac{R}{{{R}_{0}}} \right)$to be ‘y’ and the term $\ln A$to be ‘x’. We can write the new equation as follows:
$\begin{align}
  & \Rightarrow y=\dfrac{1}{3}x \\
 & \therefore y=\dfrac{x}{3} \\
\end{align}$
Which represents the equation of a straight line passing through the origin and having a slope of $\dfrac{1}{3}$ .
Hence, the graph of $\ln \left( \dfrac{R}{{{R}_{0}}} \right)$ versus $\ln (A)$of nucleus is a straight line passing through the origin.

So, the correct answer is “Option A”.

Note: The equation used by us to define the relation between the radius of an atom and its atomic number is a general equation, which is used only in solving problems. It gives us a general idea of the radius of the atom. The actual radius is influenced by a number of factors, and its equation is far more complex to derive and generalize for all the different elements.