A 0.50g sample of rock was found to have `2.5 xx 10^(-6)` mol of `._(19)^(40)K (t_(1//2)=1.3 xx 10^(9) "yr")` and `7.5 xx 10^(-6)` mol of `._(20)^(40)Ca` . How old is the rock?

Options

A. `6.5 xx 10^(8)` yr
B. `1.3 xx 10^(9)` yr
C. `2.6 xx 10^(9)` yr
D. `5.2 xx 10^(9)` yr

Answer

To determine the age of the rock based on the given amounts of potassium-40 and calcium-40, we can follow these steps:

### Step 1: Identify the initial and present amounts of potassium-40 and calcium-40.
- Present amount of potassium-40, \( N_t = 2.5 \times 10^{-6} \) mol
- Present amount of calcium-40, \( N_{Ca} = 7.5 \times 10^{-6} \) mol

### Step 2: Calculate the initial amount of potassium-40 before decay.
The decay of potassium-40 produces calcium-40. Therefore, the initial amount of potassium-40 can be calculated as:
\[ N_0 = N_t + N_{Ca} = (2.5 \times 10^{-6}) + (7.5 \times 10^{-6}) = 10.0 \times 10^{-6} \text{ mol} \]

### Step 3: Determine the decay constant (\( \lambda \)).
The decay constant can be calculated using the half-life formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given \( t_{1/2} = 1.3 \times 10^9 \) years, we can substitute this value:
\[ \lambda = \frac{\ln(2)}{1.3 \times 10^9} \approx \frac{0.693}{1.3 \times 10^9} \approx 5.34 \times 10^{-10} \text{ yr}^{-1} \]

### Step 4: Use the age formula to calculate the age of the rock.
The age of the rock can be calculated using the formula:
\[ t = \frac{1}{\lambda} \ln\left(\frac{N_0}{N_t}\right) \]
Substituting the values we have:
\[ t = \frac{1}{5.34 \times 10^{-10}} \ln\left(\frac{10.0 \times 10^{-6}}{2.5 \times 10^{-6}}\right) \]
Calculating the logarithm:
\[ \ln\left(\frac{10.0 \times 10^{-6}}{2.5 \times 10^{-6}}\right) = \ln(4) \approx 1.386 \]
Now substituting back into the age formula:
\[ t = \frac{1}{5.34 \times 10^{-10}} \times 1.386 \approx 2.6 \times 10^9 \text{ years} \]

### Final Answer:
The age of the rock is approximately \( 2.6 \times 10^9 \) years.

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