TS EAMCET · Physics · Nuclear Physics
Consider a radioactive nuclide which follow decay rate given by \(A(t)=A_0 2^{\left(t / t_0\right\}}\), where \(A(t)\) is the fraction of radioactive material remaining after time \(t\) from the initial \(A_0\) at zero time. Let \(A_1\) be the fraction of original activity which remains after 120 hours. Likewise \(A_2\) is the fraction of original activity remaining after 200 hours. If \(\frac{A_2}{A_2}=1.6\), then the half-life \(\left(t_0\right)\) will be
- A 10 hours
- B 20 hours
- C 40 hours
- D 60 hours
Answer & Solution
Correct Answer
(B) 20 hours
Step-by-step Solution
Detailed explanation
Given, decay rate, \(A=A_0 2^{-\left(\frac{t}{t_0}\right)}\) Let \(\frac{t}{t_0}=n\) where, \(t_0=\) half life. As,…
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