Respuesta :
Answer : The half-life of this radioisotope is, [tex]9\text{ years}[/tex]
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
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
[tex]k=\frac{2.303}{18\text{ years}}\log\frac{100g}{25g}[/tex]
[tex]k=0.077\text{ years}^{-1}[/tex]
Now we have to calculate the half-life, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=\frac{0.693}{0.077\text{ years}}[/tex]
[tex]t_{1/2}=9\text{ years}[/tex]
Therefore, the half-life of this radioisotope is, [tex]9\text{ years}[/tex]
