A rock is 1.5 x 10^9 yr old. The rock contains 238U which disintegrates to form 206Pb. Assume that t

Q: Why is the number of Lead nuclei equal to $N_0 - N_U$?

Because Uranium-238 decays into Lead-206 as the final stable product. Every Uranium nucleus that decays eventually becomes one Lead nucleus.

Q: What does $n = 1/3$ mean physically?

It means that the elapsed time is exactly one-third of the time required for half of the sample to decay.

Q: Can we use the exponential decay formula $N = N_0 e^{-\lambda t}$ instead?

Yes, absolutely! The formula $N = N_0(1/2)^n$ is derived from it and is often simpler to use when dealing with half-lives directly.

Q: Why did we assume no initial Lead was present?

If initial Lead were present, we wouldn't know how much of the current Lead came from Uranium decay versus how much was already there, making the age calculation impossible without more data.

Q: Is the ratio constant over time?

No, as time passes, more Uranium decays into Lead, so the ratio of Uranium to Lead continuously decreases.

Question

A rock is $1.5 	imes 10^9$ yr old. The rock contains $^{238}	ext{U}$ which disintegrates to form $^{206}	ext{Pb}$. Assume that there was no $^{206}	ext{Pb}$ in the rock initially and it is the only stable product formed by the decay. Calculate the ratio of number of nuclei of $^{238}	ext{U}$ to that of $^{206}	ext{Pb}$ in the rock. Half-life of $^{238}	ext{U}$ is $4.5 	imes 10^9$ yr. ($2^{1/3} = 1.259$)

Sigma Percentile · Accepted Answer

### The Concept of Radioactive Dating

Imagine you are holding a rock that has been around for over a billion years. How do we know its age? The secret lies locked inside its atoms.

Radioactive dating uses the steady, predictable decay of unstable isotopes to measure time. In this problem, we are looking at Uranium-238, which slowly transforms into stable Lead-206.

By comparing the amount of Uranium left to the amount of Lead formed, we can act like cosmic detectives and calculate the exact age of the rock.

### Setting Up the Decay Equation

Let's start by defining our variables. We know the age of the rock is $t = 1.5 	imes 10^9 	ext{ yr}$, and the half-life of Uranium-238 is $t_{1/2} = 4.5 	imes 10^9 	ext{ yr}$.

The first step is to figure out how many half-lives have actually passed. We do this by dividing the total time by the half-life.

$$n = \frac{t}{t_{1/2}} = \frac{1.5 	imes 10^9}{4.5 	imes 10^9} = \frac{1}{3}$$

This tells us that exactly one-third of a half-life has elapsed.

### The Mathematics of Half-Lives

Now, let's use the fundamental law of radioactive decay. If we start with an initial number of Uranium nuclei, $N_0$, the number of nuclei remaining after $n$ half-lives is given by:

$$N_U = N_0 \left(\frac{1}{2}ight)^n = N_0 \left(\frac{1}{2}ight)^{1/3}$$

But what happens to the Uranium that decays? It doesn't just vanish; it turns into Lead-206!

Since we assume there was no Lead initially, every single Lead nucleus present today came from a decayed Uranium nucleus. Therefore, the number of Lead nuclei is simply the initial Uranium minus the remaining Uranium.

$$N_{Pb} = N_0 - N_U = N_0 \left[1 - \left(\frac{1}{2}ight)^{1/3}ight]$$

### Finding the Ratio

We are asked to find the ratio of the number of Uranium nuclei to the number of Lead nuclei currently in the rock. Let's set up the ratio:

$$\frac{N_U}{N_{Pb}} = \frac{N_0 \left(\frac{1}{2}ight)^{1/3}}{N_0 \left[1 - \left(\frac{1}{2}ight)^{1/3}ight]}$$

Notice how beautifully the initial amount $N_0$ cancels out! This is why radioactive dating works regardless of the initial mass of the sample.

$$\frac{N_U}{N_{Pb}} = \frac{\left(\frac{1}{2}ight)^{1/3}}{1 - \left(\frac{1}{2}ight)^{1/3}}$$

To make the calculation easier, let's multiply the numerator and the denominator by $2^{1/3}$.

$$\frac{N_U}{N_{Pb}} = \frac{1}{2^{1/3} - 1}$$

### Final Calculation

We are given the value of $2^{1/3}$ as $1.259$. Let's substitute this into our simplified expression.

$$\frac{N_U}{N_{Pb}} = \frac{1}{1.259 - 1} = \frac{1}{0.259}$$

Performing this final division gives us our answer.

**$\frac{N_U}{N_{Pb}} \approx 3.861$**

This means for every Lead atom in the rock, there are currently about 3.86 Uranium atoms remaining.

Enter Numerical Value:

Visualized Solution

$Given Data$

$Number of Half-lives (n)$

$Remaining Uranium Nuclei (N_{U})$

$Formed Lead Nuclei (N_{P b})$

$Ratio \frac{N _{U}}{N _{P b}}$

$Final Calculation$

The Sigma Insight: Radioactivity

The Concept of Radioactive Dating

Setting Up the Decay Equation

The Mathematics of Half-Lives

Finding the Ratio

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Half-lives (n)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Remaining Uranium Nuclei (NU​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formed Lead Nuclei (NPb​)

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

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