Answer : The amount left of leutium-176 will be, 2.10 g
Solution :
First we have to calculate the rate constant, we use the formula :
\(k=\frac{0.693}{t_{1/2}}\)
\(k=\frac{0.693}{3.85\times 10^{10}\text{years}}\)
\(k=0.18\times 10^{-10}\text{years}^{-1}\)
Now we have to calculate the amount left of the sample.
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 = \(0.18\times 10^{-10}\text{years}^{-1}\)
t = decay time = \(1.155\times 10^{11}\text{ years}\)
a = initial amount of the sample = 16.8 g
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
\(1.155\times 10^{11}\text{years}=\frac{2.303}{0.18\times 10^{-10}\text{years}^{-1}}\log\frac{16.8}{a-x}\)
\(a-x=2.10g\)
Therefore, the amount left of leutium-176 will be, 2.10 g
