Question

The plot of ${\log }_{e}R(R$ is radius of nucleus) $vs{\log }_{e}A(A$ is mass no.) gives a straight line with slope equal to :

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. 3

Answer 1

Answer: (B)

$\frac{1}{3}$

As we know,
$r=r_0{A}^{1/3}$

$logr=log{r}_0+\left(1/3\right)logA$
So, slope of this line is 1/3.