Question

The half-life of radioactive isotope Zn$^{65}$ is 245 days. Find the time after which activity of Zn sample remains 75% of its initial value.

Hint:

When percentage of activity is given, always convert it into a fraction before applying decay equations.

Answer 1

Correct Answer: 102

Step 1: Write the decay constant in terms of half-life.
The relation between decay constant $K$ and half-life $t_{1/2}$ is:
\[ t_{1/2} = \frac{\ln 2}{K} \] \[ K = \frac{\ln 2}{245} \]
Step 2: Use radioactive decay law.
\[ t = \frac{1}{K} \ln \left( \frac{a_0}{a_t} \right) \]
Given that activity remains 75%,
\[ a_t = 0.75 a_0 \]
Step 3: Substitute values.
\[ t_{75%} = \frac{1}{K} \ln \left( \frac{4}{3} \right) \]
Step 4: Replace value of $K$.
\[ t_{75%} = \frac{245}{\ln 2} \ln \left( \frac{4}{3} \right) \]
Step 5: Numerical calculation.
\[ \ln \left( \frac{4}{3} \right) = \ln 4 - \ln 3 \] \[ = (2 \times 0.3010) - 0.4771 = 0.1249 \]
\[ t_{75%} = 245 \times \frac{0.1249}{0.3010} = 101.66 \approx 102 \text{ days} \]