Question

The half-life of $^{65}$Zn is 245 days. After $x$ days, 75% of the original activity remained. The value of $x$ in days is __________ (Nearest integer).}
(Given: $\log 3 = 0.4771$ and $\log 2 = 0.3010$)

Hint:

If less than one half-life has passed, the remaining activity will be greater than 50%.

Answer 1

Correct Answer: 102

Step 1: Using radioactive decay law.
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
Here, remaining activity = 75% = 0.75, half-life $t_{1/2} = 245$ days, and time = $x$ days.
Step 2: Substituting values.
\[ 0.75 = \left(\frac{1}{2}\right)^{\frac{x}{245}} \]
Taking logarithm on both sides:
\[ \log 0.75 = \frac{x}{245}\log \left(\frac{1}{2}\right) \]
Step 3: Simplifying logarithms.
\[ \log 0.75 = \log \left(\frac{3}{4}\right) = \log 3 - \log 4 \]
\[ = 0.4771 - 2(0.3010) = -0.1249 \]
Also,
\[ \log \left(\frac{1}{2}\right) = -\log 2 = -0.3010 \]
Step 4: Solving for $x$.
\[ \frac{x}{245} = \frac{-0.1249}{-0.3010} = 0.415 \]
\[ x = 245 \times 0.415 = 101.7 \approx 102 \]
Step 5: Conclusion.
The value of $x$ is approximately 102 days.