- #1

Noreturn

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## Homework Statement

## 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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- Thread starter Noreturn
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- #1

Noreturn

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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.

- #2

- 12,574

- 5,720

- #3

Noreturn

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

- #4

TSny

Homework Helper

Gold Member

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Watch units. A bequerel is defined how?

I don't follow your conversion of kg to μg.

I don't follow your conversion of kg to μg.

- #5

Noreturn

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4402 * (10^(-18)) kg = 4.40200 × 10-6 micrograms

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

- #6

- #7

TSny

Homework Helper

Gold Member

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I asked about the bequerel because your λ is in terms of years.BUT bequerel is s^-1 @ 10^-6 so answer is 4.4micrograms?

- #8

Noreturn

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- #9

- 12,574

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- #10

Noreturn

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Do I need to divide by Avogadro constant?

- #11

- 12,574

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

- #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

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:

- #13

- 12,574

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- #14

Noreturn

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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

- 12,574

- 5,720

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.The Initial should be bigger.

I prefer to look at it this way: If you have N atoms of atomic weight AW, the mass of the sample is given byJust realized I may have had it right but I forgot to convert the kg to g. So my answer should have been 6.28ug

$$m=AW \left(\frac{grams}{mole} \right)\times N (atoms) \times \frac{1}{N_{Avog.}} \left(\frac{mole}{atoms} \right)=AW\times \frac{N}{N_{Avog.}}(grams)$$

It's not a coincidence that the inverse of Avogadro's number (1.66×10

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