Fraction of Radioisotope Nuclei Decayed in Given Time Duration

To find the percentage of the initial quantity of nuclei that undergo radioactive decay in the first $t = 80\text{ s}$, we follow these steps:

1. Determine the decay constant ($\lambda$): The relationship between the half-life ($T_{1/2}$) and the decay constant ($\lambda$) is given by: $\lambda = \frac{\ln 2}{T_{1/2}}$

Given:

• $T_{1/2} = 1386\text{ s}$
• $\ln 2 = 0.693$

Substituting the given values: $\lambda = \frac{0.693}{1386}\text{ s}^{-1} = \frac{693 \times 10^{-3}}{1386}\text{ s}^{-1} = 0.5 \times 10^{-3}\text{ s}^{-1} = 5 \times 10^{-4}\text{ s}^{-1}$

2. Calculate the exponent $\lambda t$: For $t = 80\text{ s}$: $\lambda t = \left(5 \times 10^{-4}\text{ s}^{-1}\right) \times 80\text{ s} = 400 \times 10^{-4} = 0.04$

3. Calculate the fraction of nuclei that decay: The number of nuclei remaining at time $t$ is: $N(t) = N_0 e^{-\lambda t}$

The number of nuclei that undergo decay in time $t$ is: $\Delta N = N_0 - N(t) = N_0 \left(1 - e^{-\lambda t}\right)$

Thus, the fraction of nuclei decayed is: $\frac{\Delta N}{N_0} = 1 - e^{-\lambda t}$

Since $\lambda t = 0.04 \ll 1$, we can use the Taylor series expansion $e^{-x} \approx 1 - x$: $\frac{\Delta N}{N_0} \approx 1 - (1 - \lambda t) = \lambda t = 0.04$

4. Compute the percentage: $\text{Percentage decayed} = \frac{\Delta N}{N_0} \times 100\% = 0.04 \times 100\% = 4\%$

(Taking the exact value: $1 - e^{-0.04} \approx 0.0392$, which yields $3.92\%$, rounding off to the nearest integer gives $4$.)

Answer: 4