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. \[ \text{Given: }\log 2=0.30,\quad \log 3=0.48,\quad \log 4=0.60 \]

Hint:

Radioactive decay follows first order kinetics; activity is directly proportional to number of undecayed atoms.

Answer 1

Concept:
Radioactive decay follows first order kinetics. For first order decay: \[ k=\frac{0.693}{t_{1/2}} \] and \[ k=\frac{2.303}{t}\log\frac{A_0}{A} \] where \(A_0\) is initial activity and \(A\) is activity after time \(t\).

Step 1: Write the given data.
Half-life: \[ t_{1/2}=1.5\times 10^{10}\;\text{years} \] Activity becomes \(75\%\) of original value: \[ A=\frac{75}{100}A_0=\frac{3}{4}A_0 \] Therefore: \[ \frac{A_0}{A} = \frac{A_0}{\frac{3}{4}A_0} = \frac{4}{3} \]

Step 2: Use first order relation with half-life.
\[ t=\frac{2.303}{k}\log\frac{A_0}{A} \] Since: \[ k=\frac{0.693}{t_{1/2}} \] we get: \[ t=\frac{2.303t_{1/2}}{0.693}\log\frac{A_0}{A} \] Using: \[ \frac{2.303}{0.693}=3.322 \] \[ t=3.322\times t_{1/2}\times \log\frac{4}{3} \]

Step 3: Calculate \(\log \frac{4}{3}\).
\[ \log\frac{4}{3}=\log 4-\log 3 \] \[ \log\frac{4}{3}=0.60-0.48 \] \[ \log\frac{4}{3}=0.12 \]

Step 4: Calculate time.
\[ t=3.322\times 1.5\times 10^{10}\times 0.12 \] \[ t=0.598\times 10^{10} \] \[ t=5.98\times 10^9\;\text{years} \] Approximately: \[ t=6.0\times 10^9\;\text{years} \] Hence: \[ \boxed{t=6\times 10^9\;\text{years}} \]