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 :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{3.85\times 10^{10}\text{years}}[/tex]
[tex]k=0.18\times 10^{-10}\text{years}^{-1}[/tex]
Now we have to calculate the amount left of the sample.
Expression for rate law for first order kinetics is given by :
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]0.18\times 10^{-10}\text{years}^{-1}[/tex]
t = decay time = [tex]1.155\times 10^{11}\text{ years}[/tex]
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
[tex]1.155\times 10^{11}\text{years}=\frac{2.303}{0.18\times 10^{-10}\text{years}^{-1}}\log\frac{16.8}{a-x}[/tex]
[tex]a-x=2.10g[/tex]
Therefore, the amount left of leutium-176 will be, 2.10 g
