# Question a57d5

Apr 20, 2017

$\textsf{1.118 \textcolor{w h i t e}{x} g}$

#### Explanation:

As the U 238 decays exponentially, the amount of Pb 206 grows correspondingly:

The half - life of U 238 is about 4.5 billion years. As time passes, the ratio of Pb 206 to U 238 will increase.

Radioactive decay is a first order process:

$\textsf{{U}_{t} = {U}_{0} {e}^{- \lambda \text{t}}}$

$\textsf{{U}_{0}}$ is the number of undecayed U 238 atoms initially.

$\textsf{{U}_{t}}$ is the number of undecayed U 238 after time $\textsf{t}$.

$\textsf{\lambda}$ is the decay constant.

Since the decay of 1 U 238 atom will result in the formation of 1 atom of Pb 206 we can say that:

$\textsf{{U}_{0} = {U}_{t} + P {b}_{t}}$

Where $\textsf{P {b}_{t}}$ is the number of Pb 206 atoms formed after time $\textsf{t}$.

The decay equation can therefore be written:

$\textsf{{U}_{t} = \left({U}_{t} + P {b}_{t}\right) {e}^{- \lambda \text{t}}}$

$\therefore$$\textsf{\frac{{U}_{t}}{\left({U}_{t} + P {b}_{t}\right)} = {e}^{- \lambda \text{t}}}$

$\therefore$sf((U_(t)+Pb_(t))/(U_(t))=e^(lambdat)

$\therefore$$\textsf{1 + \frac{P {b}_{t}}{U} _ \left(t\right) = {e}^{\lambda \text{t}}}$

The half - life, $\textsf{{t}_{\frac{1}{2}}}$, of U 238 is $\textsf{4.47 \times {10}^{9}}$ years.

We can get the value of the decay constant from the expression:

sf(lambda=0.693/(t_(1/2))#

$\textsf{\lambda = \frac{0.693}{4.47 \times {10}^{9}} = 0.155 \times {10}^{- 9} {\text{ " "yr}}^{- 1}}$

We know that:

$\therefore$$\textsf{1 + \frac{P {b}_{t}}{U} _ \left(t\right) = {e}^{\lambda \text{t}}}$

Taking natural logs of both sides we get:

$\textsf{\ln \left[\frac{P {b}_{t}}{U} _ \left(t\right) + 1\right] = \lambda t}$

Putting in the numbers:

$\textsf{\ln \left[\frac{P {b}_{t}}{U} _ \left(t\right) + 1\right] = 0.155 \times \cancel{{10}^{-} 9} \times 4.47 \times \cancel{{10}^{9}}}$

$\textsf{\ln \left[\frac{P {b}_{t}}{U} _ \left(t\right) + 1\right] = 0.69286}$

From which:

$\textsf{\frac{P {b}_{t}}{{U}_{t}} + 1 = 1.9994}$

$\therefore$$\textsf{\frac{P {b}_{t}}{{U}_{t}} = 0.9994}$

This ratio can represent the number of moles of $\textsf{P {b}_{t} : {U}_{t}}$

It seems a reasonable result since about one half - life has elapsed so we would expect the ratio to be close to 1.

$\textsf{{U}_{t} = \frac{m}{A} _ \left(r\right) = \frac{1.305}{238} = 0.005483}$

$\therefore$$\textsf{P {b}_{t} = 0.9994 \times 0.005483 = 0.0054267}$

$\therefore$$\textsf{{m}_{P {b}_{t}} = P {b}_{t} \times {A}_{r} = 0.0054267 \times 206 = 1.118 \textcolor{w h i t e}{x} g}$