Question

Assume that ²¹⁸Po, with a half-life of 138 days, is in secular equilibrium with ²³⁸U whose half-life is 4.5 x 10⁹ y. How many grams of ²¹⁸Po will be present for each gram of ²³⁸U in the mineral? Express your answer in logarithm (to the base 10). (Round off to two decimal places)

Answer 1

Correct Answer: -10.07 - -10.08

$\text{1. Define Variables and Secular Equilibrium Condition}$

In secular equilibrium, the decay rate of the parent equals the decay rate of the daughter:

$$A_P = A_D$$

$$N_P \lambda_P = N_D \lambda_D$$

Where:

$P$ = Parent ($\text{Uranium-238, } {}^{238}\text{U}$)

$D$ = Daughter ($\text{Polonium-218, } {}^{218}\text{Po}$)

$N$ = Number of nuclei

$\lambda$ = Decay constant

The decay constant ($\lambda$) is related to the half-life ($t_{1/2}$) by:

$$\lambda = \frac{\ln(2)}{t_{1/2}}$$

Substituting the decay constants into the equilibrium equation:

$$N_P \left( \frac{\ln(2)}{t_{1/2, P}} \right) = N_D \left( \frac{\ln(2)}{t_{1/2, D}} \right)$$

This simplifies to the equilibrium ratio of the number of nuclei:

$$\frac{N_D}{N_P} = \frac{t_{1/2, D}}{t_{1/2, P}}$$

$\text{Given Values}$

$t_{1/2, D} ({}^{218}\text{Po}) = 138 \text{ days}$

$t_{1/2, P} ({}^{238}\text{U}) = 4.5 \times 10^9 \text{ years}$

$\text{2. Calculate the Mass Ratio}$

We need to find the ratio of masses ($\text{mass}_D / \text{mass}_P$). The number of nuclei ($N$) is related to mass ($m$) by:

$$N = \frac{m}{M} N_A$$

Where $M$ is the molar mass (approximately equal to the mass number in grams) and $N_A$ is Avogadro's constant.

The mass ratio is:

$$\frac{m_D}{m_P} = \frac{N_D M_D}{N_P M_P}$$

Substitute the equilibrium ratio $\left(\frac{N_D}{N_P} = \frac{t_{1/2, D}}{t_{1/2, P}}\right)$ into the mass ratio equation:

$$\frac{m_D}{m_P} = \left(\frac{t_{1/2, D}}{t_{1/2, P}}\right) \left(\frac{M_D}{M_P}\right)$$

$\text{Substitute Values}$

$\text{Mass Numbers (Molar Mass in grams):}$ $M_D = 218 \text{ g/mol}$, $M_P = 238 \text{ g/mol}$.

$\text{Half-lives (convert to the same unit, days):}$

$$t_{1/2, P} = 4.5 \times 10^9 \text{ y} \times 365.25 \frac{\text{days}}{\text{y}} = 1.643625 \times 10^{12} \text{ days}$$

$$\frac{m_D}{m_P} = \left(\frac{138 \text{ days}}{1.643625 \times 10^{12} \text{ days}}\right) \left(\frac{218 \text{ g/mol}}{238 \text{ g/mol}}\right)$$

$$\frac{m_D}{m_P} \approx (8.3963 \times 10^{-11}) \times (0.91596)$$

$$\frac{m_D}{m_P} \approx 7.6859 \times 10^{-11}$$

This value is the mass of ${}^{218}\text{Po}$ per gram of ${}^{238}\text{U}$ (since $m_P = 1 \text{ g}$).

$\text{3. Express the Answer in Logarithm}$

The question asks for the answer in $\log_{10}$ of this mass ratio:

$$\log_{10} \left(\frac{m_D}{m_P}\right) = \log_{10} (7.6859 \times 10^{-11})$$

Using the property $\log(A \times B) = \log(A) + \log(B)$:

$$\log_{10} \left(\frac{m_D}{m_P}\right) = \log_{10} (7.6859) + \log_{10} (10^{-11})$$

$$\log_{10} \left(\frac{m_D}{m_P}\right) \approx 0.8857 - 11$$

$$\log_{10} \left(\frac{m_D}{m_P}\right) \approx -10.1143$$