Question

A sample taken from a layer of mica in a canyon has 2.10 grams of potassium-40. A test reveals it to be 2.6 billion years old. How much potassium-40 was in the sample originally if the half-life of potassium-40 is 1.3 billion years? A. 4.20 g B. 8.40 g C. 12.6 g D. 16.8 g E. 25.2 g

Answer 1

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 :

$t_{1/2}=\frac{0.693}{k}$

Now put the value of half-life, we get the value of rate constant.

$1.3\times 10^9years=\frac{0.693}{k}$

$k=5.33\times 10^{-10}year^{-1}$

The expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant  = $5.33\times 10^{-10}year^{-1}$

t = time taken for decay process  = $2.6\times 10^9years$

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

$5.33\times 10^{-10}year^{-1}=\frac{2.303}{2.6\times 10^9years}\log\frac{a}{2.10g }$

$a=8.40g$

Therefore, the amount of potassium-40 in the sample originally was, 8.40 g

Answer 2

Answer:The correct answer is option B.

Explanation:

Initial amount of an isotope = $N_o$

Final amount of an isotope = N

Half life of an isotope =$t_{\frac{1}{2}}=1.3 \text{billion years}$

$\log N=\log N_o-\frac{\lambda t}{2.303}$

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{1.3 \text{billion years}}=0.5330 (\text{billion years})^{-1}$

$\log[2.10 g]=\log N_o-\frac{0.5330 (\text{billion years})^{-1}\times 2.6 \text{billion years}}{2.303}$

$0.9239=\log N_o$

$N_o=8.39266 g\approx 8.40 g$

Hence, the correct answer is option B.