Answer:
44 years.
Step-by-step explanation:
Let original amount be 100.
We have been given that the half life of Pb-210 is 22 years. A decayed animal shows 25% of the original Pb-210 remains.
We will use half-life formula to solve our given problem.
\(A=a\cdot (0.5)^{\frac{t}{h}}\), where,
A = Amount after t units of time,
a = Initial amount,
t = Time,
h = Half-life.
\(25=100\cdot (0.5)^{\frac{t}{22}}\)
\(\frac{25}{100}=\frac{100\cdot (0.5)^{\frac{t}{22}}}{100}\)
\(0.25=(0.5)^{\frac{t}{22}}\)
Take natural log of both sides:
\(\text{ln}(0.25)=\text{ln}((0.5)^{\frac{t}{22}})\)
Using natural log property \(\text{ln}(a^b)=b\cdot \text{ln}(a)\), we will get:
\(\text{ln}(0.25)=\frac{t}{22}*\text{ln}(0.5)\)
\(\frac{\text{ln}(0.25)}{\text{ln}(0.5)}=\frac{t*\text{ln}(0.5)}{22*\text{ln}(0.5)}\)
\(\frac{-1.38629436}{-0.69314718}=\frac{t}{22}\)
\(2=\frac{t}{22}\)
\(2*22=\frac{t}{22}*22\)
\(44=t\)
Therefore, the animal has been deceased to 44 years.
