Respuesta :
Answer : The correct option is, (B) 8.40 g
Solution : Given,
As we know that the radioactive decays follow the first order kinetics.
First, we have to calculate the half life of a potassium-40.
Formula used :
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
Now put the value of half-life, we get the value of rate constant.
[tex]1.3\times 10^9years=\frac{0.693}{k}[/tex]
[tex]k=5.33\times 10^{-10}year^{-1}[/tex]
The 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 = [tex]5.33\times 10^{-10}year^{-1}[/tex]
t = time taken for decay process = [tex]2.6\times 10^9years[/tex]
a = initial amount of potassium-40 = ?
a - x = amount left after decay process = 2.10 g
Now put all the values in above equation, we get
[tex]5.33\times 10^{-10}year^{-1}=\frac{2.303}{2.6\times 10^9years}\log\frac{a}{2.10g }[/tex]
[tex]a=8.40g[/tex]
Therefore, the amount of potassium-40 in the sample originally was, 8.40 g
Answer:The correct answer is option B.
Explanation:
Initial amount of an isotope = [tex]N_o[/tex]
Final amount of an isotope = N
Half life of an isotope =[tex]t_{\frac{1}{2}}=1.3 \text{billion years}[/tex]
[tex]\log N=\log N_o-\frac{\lambda t}{2.303}[/tex]
[tex]\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{1.3 \text{billion years}}=0.5330 (\text{billion years})^{-1}[/tex]
[tex]\log[2.10 g]=\log N_o-\frac{0.5330 (\text{billion years})^{-1}\times 2.6 \text{billion years}}{2.303}[/tex]
[tex]0.9239=\log N_o[/tex]
[tex]N_o=8.39266 g\approx 8.40 g[/tex]
Hence, the correct answer is option B.
