
The activity of a radioactive sample is \[1.6\;\]curie and its half-life is \[2.5\;\]days. Then activity after \[10\] days will be:
A. \[0.16\;\] curie
B. \[0.8\;\] curie
C. \[0.4\;\] curie
D. \[0.1\] curie
Answer
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Hint: In the given question, we will find the radioactivity after \[10\] days. For this, we will determine the half-life periods first. After that, we need to find the radioactivity after \[10\] days using these half-life periods and the following formula.
Formula used:
The amount of substance that will decay is given by
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^n}\]
Here, \[{N_0}\] is the amount of substance that will initially decay and \[N\] is the quantity that still remains and its decay has not taken place after a time \[t\].
Complete step by step solution:
We know that the amount of substance that will decay is,
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^n} \\ \]
So, we know that after every \[2.5\;\] days, its activity reduces to half the value. Thus,
\[10\] days \[ = \dfrac{{10}}{{2.5}} = 4\]
But, the initial activity of a radioactive sample is \[1.6\;\]curie.
So, the radioactivity after \[10\] days is given by
\[N = {N_0} \times {\left( {\dfrac{1}{2}} \right)^4} \\ \]
Here, \[{N_0} = 1.6\] curie.
This gives, \[N = 1.6 \times \left( {\dfrac{1}{4}} \right) \\ \]
By simplifying, we get
\[N = 0.1\] curie
Thus, the radioactivity after \[10\] days will be \[0.1\] curie
Therefore, the correct option is D.
Additional Information : An atom's nucleus shows radioactivity as a result of nuclear instability. Henry Becquerel made this discovery in 1896. The phenomenon of radioactivity takes place when an unstable atom's nucleus releases radiation to lose energy.
Note: Many students make mistakes in writing the formula that still remains and its decay has not taken place after a time \[t\]. Also, the value of radioactivity after \[10\] days totally depends on the half-life periods. If they make mistakes here, then the final result will be wrong. So, it is necessary to do the calculations carefully.
Formula used:
The amount of substance that will decay is given by
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^n}\]
Here, \[{N_0}\] is the amount of substance that will initially decay and \[N\] is the quantity that still remains and its decay has not taken place after a time \[t\].
Complete step by step solution:
We know that the amount of substance that will decay is,
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^n} \\ \]
So, we know that after every \[2.5\;\] days, its activity reduces to half the value. Thus,
\[10\] days \[ = \dfrac{{10}}{{2.5}} = 4\]
But, the initial activity of a radioactive sample is \[1.6\;\]curie.
So, the radioactivity after \[10\] days is given by
\[N = {N_0} \times {\left( {\dfrac{1}{2}} \right)^4} \\ \]
Here, \[{N_0} = 1.6\] curie.
This gives, \[N = 1.6 \times \left( {\dfrac{1}{4}} \right) \\ \]
By simplifying, we get
\[N = 0.1\] curie
Thus, the radioactivity after \[10\] days will be \[0.1\] curie
Therefore, the correct option is D.
Additional Information : An atom's nucleus shows radioactivity as a result of nuclear instability. Henry Becquerel made this discovery in 1896. The phenomenon of radioactivity takes place when an unstable atom's nucleus releases radiation to lose energy.
Note: Many students make mistakes in writing the formula that still remains and its decay has not taken place after a time \[t\]. Also, the value of radioactivity after \[10\] days totally depends on the half-life periods. If they make mistakes here, then the final result will be wrong. So, it is necessary to do the calculations carefully.
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