Nucleus: Practice Problem & Solution
The activity of a radioactive sample is measured as $N_0$ counts per minute at t = 0 and $N_0/e$ counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is: (2010 Pre)
Solution Explained:
To solve this problem, we apply the core principles of Nucleus. Understanding the underlying formula is key to arriving at the correct answer below:
Using $A = A_0 e^{-\lambda t}$, we get $N_0/e = N_0 e^{-5\lambda}$, yielding $5\lambda = 1$ and $\lambda = 1/5$ per minute. The half-life is $T_{1/2} = \frac{\ln 2}{\lambda} = 5 \ln 2 = 5 \log_e 2$.
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