A) \[{{A}_{1}}{{t}_{1}}={{A}_{2}}{{t}_{2}}\]
B) \[\frac{{{A}_{1}}-{{A}_{2}}}{{{t}_{2}}-{{t}_{1}}}=\text{constant}\]
C) \[{{A}_{2}}={{A}_{1}}{{e}^{\left( \frac{{{t}_{1}}-{{t}_{2}}}{T} \right)}}\]
D) \[{{A}_{2}}={{A}_{1}}{{e}^{\left( \frac{{{t}_{1}}}{T{{t}_{2}}} \right)}}\]
Correct Answer: C
Solution :
Let \[{{A}_{0}}=\]initial activity. Then, \[{{A}_{1}}={{A}_{0}}{{e}^{-\lambda {{t}_{1}}}}\]and \[{{A}_{2}}={{A}_{0}}{{e}^{-\lambda {{t}_{2}}}},\]Also, \[\lambda =\frac{1}{T}\] \[\therefore \] \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{{{e}^{-\lambda /2}}}{{{e}^{-\lambda /1}}}={{e}^{(-\lambda {{t}_{2}}+\lambda {{t}_{1}})}}\] or \[{{A}_{2}}={{A}_{1}}{{e}^{\lambda ({{t}_{1}}-{{t}_{2}})}}={{A}_{1}}{{e}^{({{t}_{1}}-{{t}_{2}})/T.}}\] Hence, the correction option is (c).
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