Question

The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reactor core spread over a wide area. The primary radiation exposure to people in western Europe was due to the short-lived (half-life 8.0 days) isotope 131I
131 I, which fell across the landscape and was ingested by grazing cows that concentrated the isotope in their milk. Farmers couldn't sell the contaminated milk, so many opted to use the milk to make cheese, aging it until the radiactivity decayed to acceptable levels. How much time must elapse for the activity of a block of cheese containing 131I 131 I to drop to 1.0% of its initial value?

Answer 1

Answer:

The correct answer is "53.15 days".

Explanation:

Given that:

Half life of $131_{I}$,

$T_{\frac{1}{2} }= 8 \ days$

• Let the initial activity be "$R_o$".
• and, activity to time t be "R".

To find t when R will be "1%" of $R_o$, then

⇒ $R=\frac{1}{100}R_o$

As we know,

⇒ $R=R_o e^{-\lambda t}$

or,

∴ $e^{\lambda t}=\frac{R_o}{R}$

By putting the values, we get

        $=\frac{R_o}{\frac{R}{100} }$

        $=100$

We know that,

Decay constant, $\lambda = \frac{ln2}{T_{\frac{1}{2} }}$

hence,

⇒ $\lambda t=ln100$

     $t=\frac{ln100}{\lambda}$

        $=\frac{ln100}{\frac{ln2}{8} }$

        $=53.15 \ days$