A) 20 min
B) 30 min
C) 40 min
D) 25 min
Correct Answer: C
Solution :
Given that the half-life of a radioactive substance is 20 min. So, \[{{t}_{1/2}}=20\,\min .\] For 20% decay, we have 80% of the substance left, hence \[\frac{80{{N}_{0}}}{100}={{N}_{0}}{{e}^{-\lambda {{t}_{20}}}}\] ?(i) where \[{{N}_{0}}=\]initial undecayed substance and \[{{t}_{20}}\] is the time taken for 20% decay. For 80% decay, we have 20% of the substance left, hence \[\frac{20{{N}_{0}}}{100}={{N}_{0}}{{e}^{-\lambda {{t}_{80}}}}\] ?(ii) Dividing Eq. (i) and Eq. (ii), we get \[4={{e}^{\lambda ({{t}_{80}}-{{t}_{20}})}}\] \[\Rightarrow \] \[\ln \,4=\lambda ({{t}_{80}}-{{t}_{20}})\] (taking log on both sides) \[\Rightarrow \] \[2\ln \,2=\frac{0.693}{{{t}_{1/2}}}({{t}_{80}}-{{t}_{20}})\] \[\Rightarrow \] \[{{t}_{80}}-{{t}_{20}}=2\times {{t}_{1/2}}\] \[=40\,\min \]
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