The isotopic composition of rubidium is `Rb^(85):73%` and `Rb^(87):28%Rb^(87)` is weakly radioactive and decay by `beta^(c-)-` emission with a decay constant of `1.1xx10^(-11) ` per year. A sample of the mineral pollucite was found to contain `450g Rb` and `0.72mg ` of `Sr^(87)`. Estimate the age of mineral pollucite.

Explanation

To estimate the age of the mineral pollucite based on the isotopic composition of rubidium and the amount of strontium present, we can follow these steps:

### Step 1: Determine the amount of Rb-87 in the sample
Given that the total amount of rubidium (Rb) in the sample is 450 g and that Rb-87 makes up 28% of the rubidium isotopes, we can calculate the amount of Rb-87.

\[ \text{Amount of Rb-87} = 450 \, \text{g} \times \frac{28}{100} = 126 \, \text{g} \]

### Step 2: Understand the decay process
Rubidium-87 decays into strontium-87 (Sr-87) through beta decay. The decay constant (\( \lambda \)) is given as \( 1.1 \times 10^{-11} \) per year.

### Step 3: Calculate the half-life of Rb-87
The half-life (\( t_{1/2} \)) can be calculated using the formula:

\[ t_{1/2} = \frac{0.693}{\lambda} \]

Substituting the decay constant:

\[ t_{1/2} = \frac{0.693}{1.1 \times 10^{-11}} \approx 6.3 \times 10^{10} \, \text{years} \]

### Step 4: Relate the amount of Rb-87 and Sr-87
Let \( A \) be the initial amount of Rb-87, and \( X \) be the amount of Rb-87 that has decayed to Sr-87. The amount of Sr-87 present is given as 0.72 mg.

From the decay relationship:

\[ A - X = \text{Current amount of Rb-87} \]

Since \( X \) is the amount that has decayed to Sr-87, we can write:

\[ X = 0.72 \, \text{mg} \]

Thus, the current amount of Rb-87 is:

\[ A - 0.72 = 126 \, \text{mg} \]

So,

\[ A = 126 + 0.72 = 126.72 \, \text{mg} \]

### Step 5: Use the decay law to find the age
The age of the sample can be calculated using the formula:

\[ t = \frac{1}{\lambda} \ln\left(\frac{A}{A - X}\right) \]

Substituting the values:

\[ t = \frac{1}{1.1 \times 10^{-11}} \ln\left(\frac{126.72}{126}\right) \]

Calculating the logarithm:

\[ \ln\left(\frac{126.72}{126}\right) \approx \ln(1.0057) \approx 0.0057 \]

Now substituting this back into the age formula:

\[ t \approx \frac{1}{1.1 \times 10^{-11}} \times 0.0057 \approx 5.09 \times 10^{8} \, \text{years} \]

### Conclusion
The estimated age of the mineral pollucite is approximately \( 5.09 \times 10^{8} \) years.

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