Answer: 43.3 days
Explanation:
Half-life of Phosphorus-32 = 14.3 days
First we have to calculate the rate constant, we use the formula :
\(k=\frac{0.693}{14.3\text{days}}\)
\(k=0.048\text{days}^{-1}\)
Now we have to calculate the days for radioactive sample to decay to 1/8 it's size
Expression for rate law for first order kinetics is given by:
\(t=\frac{2.303}{k}\log\frac{a}{a-x}\)
where,
k = rate constant = \(0.048\text{days}^{-1}\)
t = age of sample = ?
a = let initial size = 1
a - x = amount left after decay process = \(\frac{1}{8}\)
Now put all the given values in above equation, we get
\(t==\frac{2.303}{0.048}\log\frac{1}{\frac{1}{8}}\)
\(t=43.3days\)
Thus it will take 43.3 days for radioactive sample to decay to 1/8 it's size.
