# Actinium-227 has a half life of 8 x 10^3 days, decaying by a alpha-emission. Suppose that the helium gas originating from the alpha particles were collected. What volume of helium at 25°C and 748 mmHg could be obtained from 1.00 g after 100 years?

Jan 12, 2017

$V \approx 0.1 \text{ L}$

#### Explanation:

Covert $8 \times {10}^{3}$ days to years

(8000" days")/1(1" year")/(365.25" days") = 21.9" years"

Half life equation is:

$\frac{1}{2} = {e}^{\gamma t}$

Substitute 21.9 years for t:

$\frac{1}{2} = {e}^{\gamma \left(21.9 \text{ years}\right)}$

Solve for $\gamma$

$\ln \left(\frac{1}{2}\right) = \gamma \left(21.9 \text{ years}\right)$

$\gamma = - \ln \frac{2}{21.9 \text{ years}}$

Q(t) = Q_0e^(-ln(2)t/(21.9" years))

Starting with 1 gram:

Q(t) = e^(-ln(2)t/(21.9" years))

Evaluate at t = 100 years:

Q(t) = (1" g")e^(-ln(2)(100 " years")/(21.9" years))

However we actually want the difference in moles (not grams)

(1" g")/(227" g/mol")(1 - e^(-ln(2)(100 " years")/(21.9" years)))

$4.2 \times {10}^{-} 3 \text{ mol}$ has decayed creating the same number of mole of helium nuclei:

$n = 4.2 \times {10}^{-} 3 \text{ mol}$

Convert the temperature to Kelvin:

$T = 25 + 273 = 298 K$

Convert mmHg to kPa:

P = (748" mmHg")/1(0.1333" kPa")/(1" mmHg") = 99.7" kPa"

$R = 8.134 \text{ L kPa/K mol}$

$V = \frac{n R T}{P}$

$V \approx 0.1 \text{ L}$