Questions: An activity of the radioactive sample was measured 29.1 kBq and after 38 days 21.2 kBq. Define the half-life of the radioactive sample. Answer tolerancy limits have been set ± 1 days. Attention! Radioactivity decreases exponentially. Half-life is the time required for a quantity (of substance) to reduce to half of its initial value.

Solution Steps

We are given the initial activity \( A_0 = 29.1 \) kBq and the activity after 38 days \( A = 21.2 \) kBq. We need to find the half-life \( t_{1/2} \) of the radioactive sample.

The activity of a radioactive sample decreases exponentially over time and can be described by the formula: \[ A = A_0 e^{-\lambda t} \] where:

• \( A \) is the activity at time \( t \),
• \( A_0 \) is the initial activity,
• \( \lambda \) is the decay constant,
• \( t \) is the time.

We can rearrange the formula to solve for the decay constant \( \lambda \): \[ \frac{A}{A_0} = e^{-\lambda t} \] Taking the natural logarithm of both sides: \[ \ln\left(\frac{A}{A_0}\right) = -\lambda t \] Substituting the given values: \[ \ln\left(\frac{21.2}{29.1}\right) = -\lambda \cdot 38 \] \[ \ln(0.7285) = -\lambda \cdot 38 \] \[ -0.3162 = -\lambda \cdot 38 \] \[ \lambda = \frac{0.3162}{38} \] \[ \lambda = 0.0083 \, \text{days}^{-1} \]

The half-life \( t_{1/2} \) is related to the decay constant \( \lambda \) by the formula: \[ t_{1/2} = \frac{\ln(2)}{\lambda} \] Substituting the value of \( \lambda \): \[ t_{1/2} = \frac{\ln(2)}{0.0083} \] \[ t_{1/2} = \frac{0.6931}{0.0083} \] \[ t_{1/2} = 83.49 \, \text{days} \]

Final Answer