An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive materi

Q: What exactly is a half-life?

A half-life is the time required for a quantity of a radioactive substance to reduce to exactly half of its initial value. It is a constant characteristic of each specific radioactive isotope.

Q: Why do we use $(1/2)^n$ in the decay formula?

Because after 1 half-life, the amount remaining is $1/2$. After 2 half-lives, it's half of that half, which is $(1/2) \times (1/2) = (1/2)^2$. This pattern continues, so after $n$ half-lives, the fraction remaining is $(1/2)^n$.

Q: What if the required safe level was not a perfect power of 2 like 64?

You would use logarithms to solve the equation. For example, if the initial level was 50 times the safe limit, you would get $2^n = 50$, which means $n = \log_2(50)$. You would then multiply that decimal value by the half-life.

Q: Does the type of radioactive material affect the formula we use?

The exponential decay formula $R = R_0(1/2)^n$ remains exactly the same for all radioactive materials. However, the specific value of the half-life $T_{1/2}$ is unique to each isotope.

Q: Is the laboratory completely free of radiation after 108 days?

No, radioactive decay is an asymptotic process, meaning the radiation level theoretically never truly reaches absolute zero. However, after 108 days, it reaches a 'safe level' where the radiation is low enough that it no longer poses a significant health risk.

Question

An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?

Options:

(A) 64

(B) 90

(C) 108

(D) 120

Sigma Percentile · Accepted Answer

### The Radioactive Laboratory Crisis

Imagine a scenario straight out of a sci-fi movie: a nuclear laboratory has experienced an accident, resulting in the deposition of a radioactive material. The radiation level is currently $64$ times higher than what is considered safe for human operation. The material has a half-life of $18$ days. Our mission is to determine exactly how many days we must wait until the laboratory is safe to enter again.

### The Law of Radioactive Decay

To solve this, we rely on the fundamental law of radioactive decay. The activity of a radioactive sample decreases exponentially over time. After $n$ half-lives, the remaining activity $R$ is given by the formula:

$$R = R_0 \left(\frac{1}{2}ight)^n$$

Here, $R_0$ represents the initial activity of the sample. We are given that the initial activity is $64$ times the safe level, so we can write $R_0 = 64 R_{	ext{safe}}$. We want the final activity $R$ to be exactly equal to the safe level, $R_{	ext{safe}}$.

### Solving for the Number of Half-Lives

Let's substitute our known values into the decay equation:

$$R_{	ext{safe}} = 64 R_{	ext{safe}} \left(\frac{1}{2}ight)^n$$

We can easily cancel out $R_{	ext{safe}}$ from both sides of the equation, which simplifies our problem significantly:

$$1 = 64 \left(\frac{1}{2}ight)^n$$

Rearranging this equation to isolate the exponential term, we get:

$$\frac{1}{64} = \left(\frac{1}{2}ight)^n$$

Or, equivalently:

$$2^n = 64$$

We know from basic powers of $2$ that $2^6 = 64$. Therefore, comparing the exponents, we find that $n$ must be equal to $6$. This tells us that it will take exactly $6$ half-lives for the radiation to decay down to the safe level.

### Calculating the Total Time

Finally, we need to convert this number of half-lives into actual days. The total time $t$ is simply the number of half-lives $n$ multiplied by the duration of one half-life $T_{1/2}$:

$$t = n 	imes T_{1/2}$$

Substituting our values:

$$t = 6 	imes 18 	ext{ days}$$

$$t = 108 	ext{ days}$$

**The laboratory will be safe for use after exactly 108 days.**

### A Mathematical Note

What if the initial radiation was $50$ times the safe level instead of $64$? In that case, $n$ would not be a perfect integer. We would have the equation $2^n = 50$, and we would need to use logarithms to solve for $n$, specifically $n = \log_2(50)$. Always remember that logarithms are the key to unlocking exponential equations when the numbers aren't perfectly clean!

Select Answer:

Visualized Solution

$R_{0} = 64 R_{safe}$

$R = R_{0} (\frac{1}{2})^{n}$

$R_{safe} = 64 R_{safe} (\frac{1}{2})^{n}$

$1 = 64 (\frac{1}{2})^{n}$

$2^{n} = 64 ⟹ n = 6$

$t = n \times T_{1/2}$

$What if R_{0} = 50 R_{safe} ?$

The Sigma Insight: Radioactivity

The Radioactive Laboratory Crisis

The Law of Radioactive Decay

Solving for the Number of Half-Lives

Calculating the Total Time

A Mathematical Note

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Select Answer:

Visualized Solution

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