Search any question & find its solution
Question:
Answered & Verified by Expert
If decay constant of a radioactive sample is 0.05 per year, then find the time (in year) for which sample will decay by $75 \%$.
Options:
Solution:
1844 Upvotes
Verified Answer
The correct answer is:
$28$
Here, $\lambda=0.05$ per year
$\begin{aligned} & N=N_0-\frac{3 N_0}{4}=N_0 / 4, t=? \\ & \log _e \frac{N}{N_0}=\lambda t \\ & t=\frac{\log _e \frac{N_0}{N}}{\lambda}=\frac{\log _e 4}{0.05}=27.72 \text { year } \approx 28 \text { year }\end{aligned}$
$\begin{aligned} & N=N_0-\frac{3 N_0}{4}=N_0 / 4, t=? \\ & \log _e \frac{N}{N_0}=\lambda t \\ & t=\frac{\log _e \frac{N_0}{N}}{\lambda}=\frac{\log _e 4}{0.05}=27.72 \text { year } \approx 28 \text { year }\end{aligned}$
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.