The activity of a radioactive sample is measured as $N_0$ counts per minute at $t=0$ and $N_0/e$ counts per mi

Question

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 $t=5$ min. The time (in minute) at which the activity reduces to half its value is:

Accepted Answer

1. **Decay Law:** The activity $A$ of a radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$, where $\lambda$ is the decay constant .
2. **Determine Decay Constant ($\lambda$):**
   - Given: At $t = 5$ min, $A = N_0/e$ (where $A_0 = N_0$).
   - Substitute into the equation: $\frac{N_0}{e} = N_0 e^{-\lambda(5)}$.
   - $e^{-1} = e^{-5\lambda}$.
   - Equating exponents: $-1 = -5\lambda \implies \lambda = \frac{1}{5} 	ext{ min}^{-1}$.
3. **Calculate Half-Life ($T_{1/2}$):**
   - The half-life is the time required for the activity to reduce to half its initial value.
   - Formula: $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{\log_e 2}{\lambda}$ .
   - Substitute $\lambda = 1/5$: $T_{1/2} = \frac{\log_e 2}{1/5} = 5 \log_e 2$ min.

Step-by-Step Solution

Practice All PHYSICS Questions