Question

What fraction of a radioactive material will get disintegrated in a period of two half-lives?
A. whole
B. half
C. one-fourth
D. three-fourth

Answer 1

Hint:In the given question, we have to determine the fraction of a radioactive material if it will disintegrate in a period of two half-lives. We will use the following formula of the amount of substance decayed for this. After that, we will find the disintegrated part.

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,
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^n} \\ \]
Here, \[n = 2\]
This gives,
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^2} \\ \]
By simplifying, we get
\[\left( {\dfrac{N}{{{N_0}}}} \right) = \left( {\dfrac{1}{4}} \right) \\ \]
Hence, the disintegrated part is given by
\[1 - \left( {\dfrac{N}{{{N_0}}}} \right) = 1 - \dfrac{1}{4} \\ \]
By simplifying, we get
\[1 - \left( {\dfrac{N}{{{N_0}}}} \right) = \dfrac{3}{4}\]
Thus, three-fourths of a radioactive material will disintegrate in a period of two half-lives.

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 the disintegrated part totally depends on the ratio \[\left( {\dfrac{N}{{{N_0}}}} \right)\]. If they make mistakes here, then the final result will be wrong. So, it is necessary to do the calculations carefully.