Question

A radioisotope X with a half life $1.4 \times 10^9$ years decays to Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1 : 7. The age of the rock is

• A. 1.96 $\times \, 10^9$years
• B. 3.92 $\times \, 10^9$years
• C. 4.20 $\times \, 10^9$years
• D. 8.40 $\times \, 10^9$years

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the age of a rock sample based on the decay of a radioisotope X into a stable isotope Y. The ratio of the isotopes X and Y is given, and we are to find the age of the rock, given X's half-life. Here is the solution outlined clearly step-by-step:

1. Understand the given data:

• Half-life of isotope X, T_{1/2} = 1.4 \times 10^9 years.
• Current ratio of X to Y in the sample is 1:7, meaning for every 1 part of X, there are 7 parts of Y.

2. Use the decay formula:

We know that the decay of a radioactive isotope follows the formula:

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

Where N_t is the remaining quantity of the radioisotope X after time t, and N_0 is the initial quantity of X.

3. Set up the equations:

The given ratio of X to Y is 1:7, meaning only 1 part of X is left to 7 parts formed as Y:

\frac{N_t}{Y} = \frac{1}{7}

Since Y equals the amount of X that has decayed, we have:

N_0 - N_t = 7N_t

This simplifies to N_0 = 8N_t.

4. Substitute and solve for time t:

Substitute N_0 = 8N_t into the decay formula:

N_t = 8N_t \left(\frac{1}{2}\right)^{t/T_{1/2}}

Cancel N_t:

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

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

Since \frac{1}{8} = \left(\frac{1}{2}\right)^3, we equate the exponents:

\frac{t}{T_{1/2}} = 3

t = 3 \times T_{1/2}

5. Substitute for T_{1/2}:

t = 3 \times 1.4 \times 10^9 years

t = 4.2 \times 10^9 years

The age of the rock is 4.20 \times 10^9 years.