Answer
Verified
413.1k+ views
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.
$ \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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE