Question

A radio isotope X has a half-life of $1.4\times {10}^9$ yr and 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. $1.96\times {10}^9$
  
• B. $3.92\times {10}^9$
  
• C. $4.2\times {10}^{9}yr$
  
• D. $8.4\times {10}^{9}yr$

Answer 1

The correct option is C $4.2\times {10}^{9}yr$


$\frac{X}{Y}=\frac{1}{7}$
In such question always add numerator and denominator to get initial number of atoms. $X+Y=8=N_0$ before decay.
$\frac{N}{N_0}=(\frac{1}{2}{)}^n\frac{1}{8}=(\frac{1}{2}{)}^n\left(\frac{1}{2}\right)=(\frac{1}{2}{)}^{n}n=3n=\frac{T}{t_{\frac{1}{3}}}\Rightarrow T=n\times t_{\frac{1}{2}}$
$T=3\times 1.4\times {10}^9$
$T=4.2\times {10}^9$ year