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A radioactive Nucleus X converts into a stable Nucleus Y. Half-life of X is 50 year. Calculate the age of the radioactive sample when the ratio of X to Y is 1:15.

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Last updated date: 22nd Jul 2024
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Answer
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Hint: This can be solved by laws of radioactive disintegration. The no. of atoms disintegrated per second at any instant is given by radioactive decay law.
 $ \text{N}={{\text{N}}_{\text{0}}}{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $
Here, N0 is No. of atoms originally present.
N is no. of atoms left undecayed in sample t
 $ \lambda $ is decay constant.
Relation between half-life and decay constant is,
 $ \text{T}=\dfrac{\text{0}\cdot \text{6931}}{\text{ }\!\!\lambda\!\!\text{ }} $ , This is used to calculate decay constant.

Complete Step By Step Solution
We have given Nucleus X whose initial value is $ {{\text{X}}_{0}} $ . Since it converts into Y nucleus after decay,
Ratio of X and Y element is given by $ \dfrac{1}{15} $
Now, $ \text{X}+\text{Y}={{\text{Y}}_{0}} $
 $ \text{X}+\text{15X }=\text{ }{{\text{X}}_{0}} $
 $ \text{16X}={{\text{X}}_{0}} $
 $ \dfrac{{{\text{X}}_{0}}}{\text{X}}=16 $
Or, $ \dfrac{{{\text{N}}_{0}}}{\text{N}}=16 $ where $ {{\text{N}}_{0}} $ and N is the No. of atoms present initially and after decay.
Use radioactive decay law,
 $ \text{N}={{\text{N}}_{\text{0}}}{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $ --------(1)
 $ \text{ }\!\!\lambda\!\!\text{ } $ is decay constant.
t is age of sample
Now, calculate the decay constant from the
Half-life of X = 50 years.
 $ \text{ }\!\!\lambda\!\!\text{ }=\dfrac{0\cdot 693}{50} $
From eq. (1)
 $ \dfrac{\text{N}}{{{\text{N}}_{\text{0}}}}={{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $
In $ \dfrac{{{\text{N}}_{\text{0}}}}{\text{N}}=\text{ }\!\!\lambda\!\!\text{ t} $
 $ 2\cdot 303\log \dfrac{\text{N}}{{{\text{N}}_{\text{0}}}}=\text{ }\!\!\lambda\!\!\text{ t} $
Use the value of $ \dfrac{{{\text{N}}_{\text{0}}}}{\text{N}} $ and $ \lambda $ in above eq.
 $ 2\cdot 303\log 16=\dfrac{0\cdot 693}{50}\text{t} $
 $ \text{t}=\dfrac{50}{0\cdot 693}\times 2\cdot 303\times \log {{2}^{4}} $
 $ t=\dfrac{50}{0\cdot 693}\times 2\cdot 303\times 4\log 2 $
 $ \text{ }\!\![\!\!\text{ }\therefore \text{log}{{\text{n}}^{\text{m}}}\text{=m logn }\!\!]\!\!\text{ } $
 $ \text{t}=200 $ year
This is the required result.

Note
We can also use another formula, $ \dfrac{\text{N}}{{{\text{N}}_{\text{0}}}}={{\left( \dfrac{1}{2} \right)}^{^{\dfrac{\text{t}}{\text{T}}}}} $
Here, T is half-life
t is the age of radioactive samples.
Use, $ \dfrac{\text{N}}{{{\text{N}}_{\text{0}}}}\text{=}\left( \dfrac{\text{1}}{\text{16}} \right) $ , T = 50 year. (given in question).
 $ \dfrac{1}{16}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{50}}} $
 $ {{\left( 2 \right)}^{-4}}={{\left( 2 \right)}^{\dfrac{-t}{50}}} $
Taking antilog on both sides
 $ -4=\dfrac{-t}{50} $ .
t = 200 year.