Question

The half-life period of a radioactive element is \(1.5 \times 10^{10}\) years. Calculate the time in which the activity of the element is reduced to 75% of its original value. \[ \text{Given : } \log 2 = 0.30,\; \log 3 = 0.48,\; \log 4 = 0.60 \]

Hint:

Radioactive decay follows first-order kinetics. Activity, number of nuclei and concentration all decay according to the same first-order rate equation.

Answer 1

Step 1: Calculate the decay constant from half-life.
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.5 \times 10^{10}} = 4.62 \times 10^{-11}\,\text{year}^{-1} \]
Step 2: Set up the first-order equation for 75% activity remaining.
Activity reduced to 75% means $A = 0.75\,A_0$, so $\frac{A_0}{A} = \frac{1}{0.75} = \frac{4}{3}$. Using the integrated rate law: \[ t = \frac{2.303}{k}\log\frac{A_0}{A} = \frac{2.303}{k}\log\frac{4}{3} \]
Step 3: Evaluate using given log values and find t.
$\log\frac{4}{3} = \log 4 - \log 3 = 0.60 - 0.48 = 0.12$. Therefore: \[ t = \frac{2.303 \times 0.12}{4.62 \times 10^{-11}} = \frac{0.2764}{4.62 \times 10^{-11}} \approx 6 \times 10^9\,\text{years} \] \[ \boxed{t \approx 6 \times 10^9\,\text{years}} \]