Question

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?

Answer 1

`V ~~ 0.1" L"`

Explanation:

Covert `8xx10^3` days to years

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

Half life equation is:

`1/2 = e^(gammat)`

Substitute 21.9 years for t:

`1/2 = e^(gamma(21.9" years"))`

Solve for `gamma`

`ln(1/2) = gamma(21.9" years")`

`gamma = -ln(2)/(21.9" 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)))`

`4.2xx10^-3" mol"` has decayed creating the same number of mole of helium nuclei:

`n = 4.2xx10^-3" 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" L kPa/K mol"`

`V = (nRT)/P`

`V ~~ 0.1" L"`