Question

If r the radius and a is th mass number of a necleus then the graph of log r versus log a will be

Answer 1

#1 : We know that Relation :
$R = R_o A^{1/3} --(1)$
where R = Nuclear Radius
$R_o = 1.2 * 10^{-15} =1.2 fm$
A = Mass Number

#2 : We have,
Taking log in both sides of equation (1) :
$log(R) = log(R_o) + \frac{1}{3} log(A) ----(1)$

Hence, Graph of log(R) Vs. log(A) will be a straight line whose :
Slope : 1/3
y- intercept = $log(R_o)$