Search any question & find its solution
Question:
Answered & Verified by Expert
The half-life of ${ }^{198} \mathrm{Au}$ is 2.7 days. The average life is
Options:
Solution:
1289 Upvotes
Verified Answer
The correct answer is:
3.9 days
The half-life and the decay constant are related as
$\begin{aligned} t_{12} & =\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{2.7 \mathrm{day}} \\ & =\frac{0.693}{2.7 \times 24 \times 3600 \mathrm{~s}} \\ & =2.9 \times 10^{-6} \mathrm{~s}^{-1}\end{aligned}$
Now average life
$t_{a v}=\frac{1}{\lambda}=3.9$ days
$\begin{aligned} t_{12} & =\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{2.7 \mathrm{day}} \\ & =\frac{0.693}{2.7 \times 24 \times 3600 \mathrm{~s}} \\ & =2.9 \times 10^{-6} \mathrm{~s}^{-1}\end{aligned}$
Now average life
$t_{a v}=\frac{1}{\lambda}=3.9$ days
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.