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

Concept: The activity of a radioactive substance is directly proportional to the number of undecayed nuclei present. Radioactive decay follows first-order kinetics. The relation between activity and time is: \[ A=A_0e^{-kt} \] or \[ k=\frac{2.303}{t} \log \frac{A_0}{A} \]

Step 1: Calculate decay constant. Given: \[ t_{1/2}=1.5\times10^{10}\;\text{years} \] For radioactive decay, \[ k=\frac{0.693}{t_{1/2}} \] \[ k = \frac{0.693}{1.5\times10^{10}} \] \[ k = 4.62\times10^{-11} \; \text{year}^{-1} \]

Step 2: Determine final activity. Activity reduced to \(75\%\) of original value: \[ A=0.75A_0 \] Substituting into first-order equation: \[ k = \frac{2.303}{t} \log \frac{A_0}{0.75A_0} \] \[ k = \frac{2.303}{t} \log \frac{4}{3} \] \[ \log \frac{4}{3} = \log4-\log3 \] \[ =0.60-0.48 \] \[ =0.12 \] Hence, \[ k = \frac{2.303\times0.12}{t} \] \[ k = \frac{0.276}{t} \]

Step 3: Calculate time. \[ t = \frac{0.276}{4.62\times10^{-11}} \] \[ t = 5.97\times10^{9} \] years \[ \boxed{ t \approx 6\times10^9 \text{ years} } \]

Final Answer: \[ \boxed{ 6\times10^9\ \text{years} } \] The activity will reduce to \(75\%\) of its original value in approximately \[ \boxed{6\times10^9\ \text{years}} \]