If 2.338 g of 226Ra (t1/2 = 1620 y) is separated from its decay products and pla

Question

If 2.338 g of 226Ra (t1/2 = 1620 y) is separated from its decay products and placed in a sealed 1L vessel, what is the partial pressure (in torr) of helium after 45 days if the vessel is at 0oC? The half-life of Rn is = 3.82 d.

Do not assume that 218Po is stable. Consult your nuclear wallet card to determine what it will decay to, and what this daughter will decay to, etc. This decay chain extends to 210Pb, which has a long half-life compared to the measurement time. Hint: Despite the long decay chain, you will only need to consider three isotopes (apart from helium) in your calculation. (Why? Consider the half-lives!)

The gas constant R = 62.363 L*torr*K-1*mol-1

Explanation / Answer

Since you are going from a mass number of 226 to 218 (8) and that Ra and Po have a difference of 2 protons, I'm not sure what the decay mechanism is. I'm going to assume one alpha particle (a helium nucleus) is emitted per atom of 226Ra but you need to confirm that. If it's say 2, you need to double my numbers.

45 days / 3.82 days/half life = 11.78 half lives

0.5^n = fraction remaing

0.5^11.78 = 0.000284 so nearly all the radon decays

If you had 2.338 g initially, 2.336 g decays. 2.336 g of Ra = 0.0105 moles which is the moles of He created during the decay process

Then you just use the ideal gas equation, PV=nRT. Substitute and solve for the pressure of radon.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.