Question

The graph of $\ln \left( {R/{R_0}} \right)$ versus $\ln A$ ($R =$ radius of a nucleus and $A$= its mass number) is then
A. straight line
B. a parabola
C. an ellipse
D. none of the above

Answer 1

We know that the relation between radius and mass number of nucleus is given as
$R = {R_0}{A^{\dfrac{1}{3}}}$
Where $R$ is the radius of the nucleus, $A$ is the mass number of the nucleus and ${R_0}$ is a constant.
According to logarithm power rule $\ln {a^b} = b\ln a$.

Complete step by step answer:
Given,
$R$ is the radius of a nucleus, $A$ is the mass number of a nucleus.
We know that the relation between radius and mass number of nucleus is given as
$R = {R_0}{A^{\dfrac{1}{3}}}$
Where, ${R_0}$ is a constant.
Therefore,
$\dfrac{R}{{{R_0}}} = {A^{\dfrac{1}{3}}}$ …… (1)
Take logarithm on both sides of equation (1). We get,
$\ln \left( {\dfrac{R}{{{R_0}}}} \right) = \ln \left( {{A^{\dfrac{1}{3}}}} \right)$
Since, according to logarithm power rule $\ln {a^b} = b\ln a$, we can write
$\ln \left( {\dfrac{R}{{{R_0}}}} \right) = \dfrac{1}{3}\ln A$
This equation is of the form $y = mx$
Which is the equation of a straight line with slope $m$ .
On comparing we can see that the slope of the graph will be $\dfrac{1}{3}$.
So, if we draw the graph of this equation by taking $\ln \left( {\dfrac{R}{{{R_0}}}} \right)$ on the Y axis and $\ln A$ on the X axis.
The graph that we get will be a straight line with slope $\dfrac{1}{3}$
So, the correct answer is option (A).

Note: Formulas to remember-
The relation between radius and mass number of nucleus is given as
$R = {R_0}{A^{\dfrac{1}{3}}}$
Where $R$ is the radius of the nucleus, $A$ is the mass number of the nucleus and ${R_0}$ is a constant having a value of 1.1 fm.