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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.

Last updated date: 29th Feb 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.