NEET Solved Paper 2014 Question 43
Question: A radio isotope X with a half-life $ 1.4\times 10^{9} $ yr decays of 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 [AIPMT 2014]
Options:
A) $ 1.96\times 10^{9},yr $
B) $ 3.92\times 10^{9},yr $
C) $ 4.20\times 10^{9}yr $
D) $ 8.40\times 10^{9},yr $
Show Answer
Answer:
Correct Answer: C
Solution:
- Ratio of X : Y is given =1:7 $ \frac{m _{x}}{m _{y}}=\frac{1}{7} $
$ \Rightarrow $ $ 7m _{x}=m _{y} $ Let the initial total mass is m.
$ \Rightarrow ,m _{x}+m _{y}=m\Rightarrow \frac{m^{y}}{7}+m _{y}=m $
Þ $ \frac{8m _{y}}{7}=m $
$ \Rightarrow m _{y}=\frac{7}{8}m $ only $ \frac{1}{8} $ part remains Þ $ 1\xrightarrow{\begin{smallmatrix} {{T} _{1/2}} \\ \end{smallmatrix}},\frac{1}{2}\xrightarrow{\begin{smallmatrix} {{T} _{1/2}} \\ \end{smallmatrix}},\frac{1}{4},\xrightarrow{\begin{smallmatrix} {{T} _{1/2}} \\ \end{smallmatrix}}\frac{1}{8} $ So, time taken to become $ \frac{1}{8} $ unstable part $ =3\times {{T} _{1/2}}=3\times 1.4\times 10^{9} $ $ =4.2\times 10^{9}y $
