Answer : The half-life of this radioisotope is,

Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = ?
t = time passed by the sample = 18 years
a = initial amount of the reactant = 100 g
a - x = amount left after decay process = 25 g
Now put all the given values in above equation, we get


Now we have to calculate the half-life, we use the formula :



Therefore, the half-life of this radioisotope is,
