A radioisotope X with a half-life 1.4 × 10^9 yr decays into Y which is stable. A sample of the rock from a cav

Question

A radioisotope X with a half-life 1.4 × 10^9 yr decays into 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:

Accepted Answer

1. **Understand the Decay Process:** The radioisotope X (parent) decays into stable isotope Y (daughter). The total number of nuclei is conserved. If we assume the rock initially contained only X, then the initial amount of X ($N_0$) is equal to the sum of the varying amounts of X and Y present at time $t$.
   $$ N_0 = N_X + N_Y $$
2. **Analyze the Ratio:** The observed ratio of X to Y is given as $1:7$. This means for every 1 nucleus of X remaining, there are 7 nuclei of Y formed.
   - Fraction of X remaining ($N/N_0$) = $\frac{N_X}{N_X + N_Y} = \frac{1}{1 + 7} = \frac{1}{8}$.
3. **Calculate Number of Half-Lives ($n$):** Using the radioactive decay formula $N = N_0 \left(\frac{1}{2}ight)^n$ :
   $$ \frac{1}{8} = \left(\frac{1}{2}ight)^n $$
   $$ \left(\frac{1}{2}ight)^3 = \left(\frac{1}{2}ight)^n \implies n = 3 $$
   So, 3 half-lives have passed.
4. **Calculate Age ($t$):**
   $$ t = n 	imes T_{1/2} $$
   $$ t = 3 	imes (1.4 	imes 10^9 	ext{ yr}) $$
   $$ t = 4.2 	imes 10^9 	ext{ yr} $$

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