Question

The half life of a radioactive isotope 'X' is 20 years. It decays to another element 'Y' which is stable. The two elements 'X' and 'Y' were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be

• A. 40 years
• B. 60 years
• C. 80 years
• D. 100 years

Answer 1

The Correct Option is B

The problem involves determining the age of a rock based on the radioactive decay of an isotope 'X' into a stable element 'Y'. We know from the problem statement:

• The half-life of isotope 'X' is 20 years.
• The ratio of 'X' to 'Y' in the rock sample is 1:7.

Let's denote:

• \( N_0 \): Initial quantity of 'X' at time \( t = 0 \).
• \( N \): Quantity of 'X' remaining after time \( t \).

The total quantity of 'X' and 'Y' in the sample can be expressed as \( N_0 = N + 7N \) because 'X' would have decayed to form 'Y'. Thus:

• \( N_0 = 8N \).

The decay formula for a radioactive substance is given by:

N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}

Where \( T_{1/2} \) is the half-life of the substance. Here, \( T_{1/2} = 20 \) years.

Substituting \( N_0 = 8N \) into the decay formula:

N = 8N \left(\frac{1}{2}\right)^{\frac{t}{20}}

Solving for \( \left(\frac{1}{2}\right)^{\frac{t}{20}} \), we get:

\left(\frac{1}{2}\right)^{\frac{t}{20}} = \frac{1}{8}

Taking the logarithm on both sides, we have:

\log_{1/2} \frac{1}{8} = \frac{t}{20}

Since \( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), the above can be written as:

3 = \frac{t}{20}

Multiplying through by 20 gives:

t = 60 \text{ years}

Thus, the age of the rock is 60 years.

Therefore, the correct answer is 60 years.