Radioactive Decay Calculation for ${}^{65}\text{Zn}$

The relationship between the remaining activity $A(t)$, the initial activity $A_0$, the time elapsed $t$, and the half-life $t_{1/2}$ is given by the formula:

$ \frac{A(t)}{A_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} $

In this problem:

• Initial Activity = $A_0$
• Remaining Activity = 75% of $A_0$, so $A(t) = 0.75 A_0$
• Half-life $t_{1/2}$ = 245 days
• Time elapsed $t$ = $x$ days

Substitute the known values into the formula:

$ 0.75 = \left(\frac{1}{2}\right)^{\frac{x}{245}} $

Rewrite 0.75 as a fraction:

$ \frac{3}{4} = \left(\frac{1}{2}\right)^{\frac{x}{245}} $

To solve for $x$, take the base-10 logarithm of both sides:

$ \log\left(\frac{3}{4}\right) = \log\left(\left(\frac{1}{2}\right)^{\frac{x}{245}}\right) $

Using logarithm properties ($\log(a/b) = \log a - \log b$ and $\log(a^b) = b \log a$):

$ \log 3 - \log 4 = \frac{x}{245} \log\left(\frac{1}{2}\right) $

Further simplify using $\log 4 = \log(2^2) = 2 \log 2$ and $\log(1/2) = \log 1 - \log 2 = 0 - \log 2 = -\log 2$:

$ \log 3 - 2 \log 2 = \frac{x}{245} (-\log 2) $

Now, substitute the given logarithm values ($\log 3 = 0.4771$ and $\log 2 = 0.3010$):

$ 0.4771 - 2(0.3010) = \frac{x}{245} (-0.3010) $

Calculate the values:

$ 0.4771 - 0.6020 = -\frac{0.3010 \cdot x}{245} $

$ -0.1249 = -\frac{0.3010 \cdot x}{245} $

Solve for $x$:

$ x = \frac{0.1249 \times 245}{0.3010} $

$ x \approx \frac{30.5005}{0.3010} \approx 101.33 $

Rounding to the nearest integer, the value of $x$ is 102 days.