Animated Solution for Chemistry - Chemical Kinetics: A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?

Radioactive Decay

Imagine a room where a radioactive element has just spilled. The activity right now is ten times the safe limit. We need to find out how long we have to wait until it is safe to enter. We are given that the half-life of the element is $30$ days. This is a classic problem of chemical kinetics, specifically dealing with first-order reactions.

Radioactive decay always follows first-order kinetics. This means the rate at which the element decays is directly proportional to the amount of the element currently present. A very handy formula for this is that the final activity $A$ equals the initial activity $A_0$ times one-half raised to the power of $n$, where $n$ is the number of half-lives that have passed.

Let us substitute our values. The final activity we want is the safe limit, which we can call $A_{\text{safe}}$, and the initial activity is ten times that, so $A_0=10A_{\text{safe}}$.

The safe limit cancels out on both sides, leaving us with:

Rearranging this, we get:

Now, how do we solve for $n$? We take the common logarithm (base 10) on both sides. This gives us:

Using the power rule of logarithms, we bring the $n$ down:

Since ${\log }_{10}10$ is $1$ and ${\log }_{10}2$ is approximately $0.301$, $n$ comes out to be:

This means it takes about $3.32$ half-lives for the activity to drop to a safe level.

Finally, to find the total time $t$, we multiply the number of half-lives $n$ by the half-life duration $t_{1/2}$.

Multiplying $3.32$ by $30$ days gives us $99.6$ days. Rounding this off, we get approximately $100$ days. So, it will be safe to enter the room after a hundred days.

Alternate Method: You could also solve this using the standard first-order rate equation. First, find the decay constant $k$ using $k=\frac{0.693}{t_{1/2}}$. Then, use the time equation $t=\frac{2.303}{k}\log \left(\frac{A_0}{A}\right)$. Both methods are perfectly valid and will give you the exact same result!