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(D) The fraction of radioactive material remaining after $n$ half-lives is given by the formula $\frac{N}{N_0} = \left( \frac{1}{2} \right)^n$.Given t

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Three-fourth

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(D) The fraction of radioactive material remaining after $n$ half-lives is given by the formula $\frac{N}{N_0} = \left( \frac{1}{2} \right)^n$.Given that the time period is $n = 2$ half-lives.Substituting the value of $n$,we get $\frac{N}{N_0} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.This represents the fraction of the material that remains undecayed.The fraction of the material that has disintegrated is given by $1 - \frac{N}{N_0}$.Therefore,the disintegrated fraction $= 1 - \frac{1}{4} = \frac{3}{4}$.

What fraction of a radioactive material will get disintegrated in a period of two half-lives?

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