# Find Mass given half life

Noreturn

## Homework Statement

http://d2vlcm61l7u1fs.cloudfront.net/media/f62/f62b6919-e60f-405b-81fb-2323ceeb03ee/phpsL4bVl.png

## Homework Equations

So I need that in micrograms tho. So 4402*10^-18/1000=4.4*10^-18kg. or 4.4*10^-12 micrograms

that stills say it's wrong tho.

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Homework Helper
Gold Member
Perhaps you should use the activity equation A(t) = A0 e-λt to find the initial activity A0 and convert that to the initial number of nuclei.

Noreturn
Didn't we do that at the bottom of the image?

Homework Helper
Gold Member
Watch units. A bequerel is defined how?

Noreturn
Oh! So:
4402 * (10^(-18)) kg = 4.40200 × 10-6 micrograms

BUT bequerel is s^-1 @ 10^-6 so answer is 4.4micrograms?

Homework Helper
Gold Member
Didn't we do that at the bottom of the image?
Not really. To reinforce what @TSny posted, assuming that you found A0, what is the number for λ in N0 = λ A0 when A0 has Bq units?

Homework Helper
Gold Member
BUT bequerel is s^-1 @ 10^-6 so answer is 4.4micrograms?

Noreturn
I was not able to figure this out. Any help on where I went wrong is appreciated. So I'm guessing I did the -.1322 wrong based on what you guys have mentioned.

Homework Helper
Gold Member
Look at the equation A = λ N. Put in the numbers including the units attached to each number. Then you will see what is going on.

Noreturn
Do I need to divide by Avogadro constant?

Homework Helper
Gold Member
Do I need to divide by Avogadro constant?
Not in this equation. Just do what I suggested.

Noreturn
4*10^9 Bq =.132yr^-1*N

or

4*10^9Bq = 4.18291693 × 10-9 Bq * N

N= 9.56*10^17

9.56*19^17/e^(.132*3) = 6.43*10^17

(6.43*10^17)(1.66*10^-27)(58.93) =6.28*10^-8kg or 6.3ug

Last edited:
Homework Helper
Gold Member
That's better but still incorrect. What does N= 9.56*10^17 represent? What about 6.43*10^17? What number is that? If N is the number of undecayed nuclei, after 3 years (when the activity is 4*10^9Bq) and 6.43*10^17 is the initial number of undecayed nuclei, which one should be the larger number? Also, what are the units of 1.66*10^-27 and 58.93?

Noreturn
So Muliplying 1.66*10^-27 and 58.93 converts mass number of Cobalt from amu to kg.

The Initial should be bigger.

Just realized I may have had it right but I forgot to convert the kg to g. So my answer should have been 6.28ug

Homework Helper
Gold Member
The Initial should be bigger.
I agree, but when you start with the equation A = λ N, then you say A = 4*10^9Bq, that "A" is the final A and the number N = 9.56*10^17 that you get from it is the final N. So, what is the initial N that should be larger than 9.56*10^17? That's why I suggested that you find the initial activity in post #2.
Just realized I may have had it right but I forgot to convert the kg to g. So my answer should have been 6.28ug
I prefer to look at it this way: If you have N atoms of atomic weight AW, the mass of the sample is given by
$$m=AW \left(\frac{grams}{mole} \right)\times N (atoms) \times \frac{1}{N_{Avog.}} \left(\frac{mole}{atoms} \right)=AW\times \frac{N}{N_{Avog.}}(grams)$$
It's not a coincidence that the inverse of Avogadro's number (1.66×10-24) matches the mass of a proton (or neutron) in grams, 1.67×10-24.