Question

A moon rock collected by a U.S. Apollo mission is estimated to be 4.20 billion years old by uranium/lead dating. Assuming that the rock did not contain any lead when it was formed, what is the current mass of 206Pb per 1.305 g of 238U in the rock? The half-life of 238U is t1/2 = 4.47 × 109 years.

Answer 1

Answer : The current mass of 206-Pb per 1.305 g of 238-U in the rock is, 0.54 grams.

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

To calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{4.47\times 10^9\text{ years}}$

$k=1.55\times 10^{-10}\text{ years}^{-1}$

Now we have to calculate the moles of 238-U.

$\text{Moles of 238-U}=\frac{\text{Given mass}}{\text{Molecular mass}}=\frac{1.305g}{238g/mole}=5.48\times 10^{-3}mole$

Now we have to calculate the amount left.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process

a = initial amount or moles of the reactant

a - x = amount or moles left after decay process

Putting values in above equation, we get:

$1.55\times 10^{-10}=\frac{2.303}{4.20\times 10^9}\log\frac{5.48\times 10^{-3}}{a-x}$

$a-x=2.86\times 10^{-3}mole$

Now we have to calculate the decomposed moles of uranium.

$\text{Decomposed moles of uranium}=\text{Initial moles of uranium}-\text{Left moles of uranium}=(5.48\times 10^{-3})-(2.86\times 10^{-3})=2.62\times 10^{-3}mole$

As we know that,

1 mole of $_{92}^{238}\textrm{U}$ is produced by 1 mole $_{82}^{206}\textrm{Pb}$

The decomposed moles of Pb = $2.62\times 10^{-3}mole$

Now we have to calculate the current mass of 206-Pb per 1.305 g of 238-U in the rock.

$\text{Mass of 206-Pb}=\text{Decomposed moles of Pb}\times \text{Molar mass of Pb}=2.62\times 10^{-3}\times 206=0.54g$

Therefore, the current mass of 206-Pb per 1.305 g of 238-U in the rock is, 0.54 grams.

Answer 2

The current mass of 206Pb per 1.305g of 238U in the rock is 1.037 grams

To be able to solve this question, we need to know the rate constant and the half-life of the radioactive isotope.

What is the rate constant?

From the first order of kinetics, the rate constant is the proportionality constant in relation to the half-life.

It can be expressed by using the formula:

$\mathbf{k = \dfrac{0.693}{t_{1/2}}}$

$\mathbf{k = \dfrac{0.693}{4.47 \times 10^9 \ years}}$

$\mathbf{k =1.55 0 \times 10^{-10} }$

Now, from the parameters given:

• The mass of 238U = 1.350 g
• Suppose we assume that the mass of 206Pb = x

Therefore, the equation for the decay constant can be computed as:

$\mathbf{1.305 = (1.30 5+ x) e^{-(kt)}}$

$\mathbf{1.305 = (1.30 5+ x) e^{-(1.550\times 10^{-10} \times 4.20 \times 10^9 )}}$

$\mathbf{1.305 = (1.30 5+ x) \times 0.521524}$

$\mathbf{\dfrac{1.305}{ 0.521524} = (1.30 5+ x)}$

$\mathbf{1.30 5+ x = 2.503}$

x = 2.503 - 1.305

x = 1.198

Now, the mass of the  206Pb is calculated as:

$\mathbf{=\dfrac{1.198 \times 206}{238}}$

= 1.037 grams

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