Find Mass given half life

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  • #1
Noreturn
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Homework Statement



media%2Ff62%2Ff62b6919-e60f-405b-81fb-2323ceeb03ee%2FphpsL4bVl.png
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Homework Equations


media%2Fbf0%2Fbf0a12a7-f069-41fa-a9d3-c6a3fb1c334b%2FphpNHbZIM.png

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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Answers and Replies

  • #2
kuruman
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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.
 
  • #3
Noreturn
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Didn't we do that at the bottom of the image?
 
  • #4
TSny
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Watch units. A bequerel is defined how?

I don't follow your conversion of kg to μg.
 
  • #5
Noreturn
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Oh! So:
4402 * (10^(-18)) kg = 4.40200 × 10-6 micrograms

BUT bequerel is s^-1 @ 10^-6 so answer is 4.4micrograms?
 
  • #6
kuruman
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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?
 
  • #7
TSny
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BUT bequerel is s^-1 @ 10^-6 so answer is 4.4micrograms?
I asked about the bequerel because your λ is in terms of years.
 
  • #8
Noreturn
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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.
 
  • #9
kuruman
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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.
 
  • #10
Noreturn
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Do I need to divide by Avogadro constant?
 
  • #11
kuruman
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Do I need to divide by Avogadro constant?
Not in this equation. Just do what I suggested.
 
  • #12
Noreturn
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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
 
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  • #13
kuruman
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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?
 
  • #14
Noreturn
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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
 
  • #15
kuruman
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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.
 

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