A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 time

Q: Why do we use $(1/2)^n$ instead of $e^{-\lambda t}$?

Both formulas are perfectly valid! However, when the ratio of intensities is a clean power of 2 (like 64), using the half-life formula $(1/2)^n$ makes the calculation much faster and avoids logarithms.

Q: What if the intensity ratio wasn't a perfect power of 2?

If the ratio was, say, 50, you would have to use logarithms. You'd set up the equation $1/50 = (1/2)^n$, take the log of both sides, and solve for $n$.

Q: Does the safe level depend on the material?

Yes, the "permissible safe level" depends on the type of radiation (alpha, beta, gamma) and the specific biological impact of the isotope. But mathematically, it just acts as our target final intensity.

Q: Why does the intensity halve exactly every 2 hours?

Radioactivity is a quantum mechanical process based on probability. For this specific isotope, any given nucleus has a 50% chance of decaying every 2 hours. Macroscopically, this means the overall intensity halves.

Q: Can the intensity ever reach absolute zero?

Mathematically, an exponential decay curve never truly reaches zero; it just gets infinitely close. However, physically, once you are down to the last few undecayed atoms, the continuous math model breaks down, and eventually, the very last atom will decay.

Question

A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is

Options:

(A) 6 h

(B) 12 h

(C) 24 h

(D) 128 h

Sigma Percentile · Accepted Answer

### The Danger Zone: Understanding the Setup

Imagine you have just prepared a brand new radioactive source in the lab. It is fresh, highly active, and currently emitting radiation at an intensity that is **64 times the permissible safe level**.

Handling it right now would be extremely dangerous. We need to wait for the radioactive decay process to naturally reduce the intensity down to the safe level.

The question is: exactly how long do we have to wait? We are given that the **half-life** of this material is $2 	ext{ hours}$. This means every $2 	ext{ hours}$, the intensity of the radiation will be cut exactly in half.

### The Mathematics of Waiting: Radioactive Decay

To solve this, we rely on the fundamental law of radioactive decay. Instead of using the continuous exponential function $e^{-\lambda t}$, it is much more intuitive here to use the half-life formula.

The intensity $I$ after $n$ half-lives is given by the equation:
$$I = I_0 \left(\frac{1}{2}ight)^n$$

Here, $I_0$ is the initial intensity, and $n$ represents the total number of half-lives that have passed. The total time $t$ is simply the number of half-lives multiplied by the duration of one half-life:
$$t = n 	imes T_{1/2}$$

### Crunching the Numbers: Finding the Half-Lives

Let's substitute our known values into the decay equation. We want our final intensity $I$ to be exactly equal to the safe level, $I_{	ext{safe}}$. We also know our initial intensity $I_0$ is $64 	imes I_{	ext{safe}}$.

Plugging these in, we get:
$$I_{	ext{safe}} = 64 I_{	ext{safe}} \left(\frac{1}{2}ight)^n$$

Notice how $I_{	ext{safe}}$ appears on both sides? We can safely divide both sides by $I_{	ext{safe}}$, which beautifully simplifies our equation to:
$$1 = 64 \left(\frac{1}{2}ight)^n$$

Rearranging this to isolate the fractional term, we divide by $64$:
$$\frac{1}{64} = \left(\frac{1}{2}ight)^n$$

Now, we need to express $\frac{1}{64}$ as a power of $\frac{1}{2}$. If we recall our powers of $2$, we know that $2^6 = 64$. Therefore:
$$\left(\frac{1}{2}ight)^6 = \left(\frac{1}{2}ight)^n$$

By comparing the exponents on both sides, it becomes immediately clear that:
$$n = 6$$

This tells us that exactly **6 half-lives** must pass for the radiation to drop to a safe level.

### The Final Countdown: Calculating Total Time

We are almost at the finish line! We know we need to wait for $6$ half-lives, and the problem states that each half-life is $2 	ext{ hours}$ long.

To find the total waiting time $t$, we simply multiply the number of half-lives by the duration of one half-life:
$$t = 6 	imes 2 	ext{ hours}$$
$$t = 12 	ext{ hours}$$

So, you can safely return to the lab and work with the radioactive source after a **12-hour wait**. This problem perfectly illustrates the power of exponential decay—even a highly dangerous radiation level can become safe in a relatively short amount of time!

Animated Solution for Physics - Atoms and Nuclei: A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is

Select Answer:

Visualized Solution

$Initial Setup$

$Decay Law$

$Substitution$

$Simplification$

$Finding n$

$Total Time$

$Conclusion$

The Sigma Insight: Radioactivity

The Danger Zone

The Mathematics of Waiting

Crunching the Numbers

The Final Countdown

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Animated Solution for Physics - Atoms and Nuclei: A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decay Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Time

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion