A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ___________. (Nearest integer)
(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)
Answer (integer)
75
Solution
$\lambda=\frac{2.303}{t} \log \frac{A_{0}}{A}$
<br/><br/>
$$
\begin{aligned}
&\frac{0.693}{200}=\frac{2.303}{83} \log \frac{A_{0}}{A} \\\\
&\frac{A}{A_{0}}=0.75
\end{aligned}
$$
<br/><br/>
Hence, percentage of original activity remaining after 83 days is $75 \%$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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