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Topic: The weight in grams of 1.0 curie F-18  (Read 3960 times)

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Offline preciouspup87

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The weight in grams of 1.0 curie F-18
« 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.

Offline mjc123

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Re: The weight in grams of 1.0 curie F-18
« Reply #1 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?

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