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The half-life of a radioactive sample is \(T\). The fraction of the initial mass of the sample that decays in an interval \(T / 2\) is
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The correct answer is:
\(\frac{1}{\sqrt{2}}\)
Fraction remains after \(n\) half-lives is given as
\(\frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{t / T} \quad\) [where, \(T\) is half-life]
\(\Rightarrow \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{t}{T}}\)
Given, \(\quad t=\frac{T}{2}\)
\(\therefore \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{T / 2}{T}}=\left(\frac{1}{2}\right)^{\frac{1}{2}}=\frac{1}{\sqrt{2}}\)
\(\frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{t / T} \quad\) [where, \(T\) is half-life]
\(\Rightarrow \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{t}{T}}\)
Given, \(\quad t=\frac{T}{2}\)
\(\therefore \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{T / 2}{T}}=\left(\frac{1}{2}\right)^{\frac{1}{2}}=\frac{1}{\sqrt{2}}\)
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