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Specialty Chemistry Forums => Nuclear Chemistry and Radiochemistry Forum => Topic started by: preciouspup87 on August 17, 2023, 09:38:56 PM

Title: The weight in grams of 1.0 curie F-18
Post by: preciouspup87 on August 17, 2023, 09:38:56 PM

I'm trying to see if I'm on the right track or if I'm completely off.

I have 1.0 curie of F-18 with a half-life of 109.7 min.

I used (dN/dt)=λN

Divided both sides by λ, so I have N=(1/λ)(dN/dt)

I plug in the numbers
(dN/dt)=(3.7E10 x 60 dpm)
λ=(6.32E-3)

I can flip the lambda fraction and change the charge of the exponent and multiply.
So
(1/λ)(dN/dt) = (6.32E3 x 3.7E10 x 60)

I divide by avogadros number

(6.32E3 x 3.7E10 x 60)/(6.023E23) gives atoms so I need to multiply by the atomic mass of F-18


((6.32E3 x 3.7E10 x 60)/(6.023E23)) x (18.998) = 4.43E-7 grams

Does this look right? I haven't done radiochemistry since I graduated in 2015.

Title: Re: The weight in grams of 1.0 curie F-18
Post by: mjc123 on August 18, 2023, 05:07:08 PM

You're basically on the right track, but some points:
Why is λ = 6.23e-3? (I can see why, but it's good to make it explicit, in case mistakes creep in.)
If λ = 6.23e-3, 1/λ is not 6.23e3. What is it?
Dividing by Avogadro's number gives you moles, not atoms, and you need to multiply by the molar mass ("atomic weight"). Your procedure is correct but the wording is wrong.
What is the molar mass of F-18? Is it 18.998?