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April 16, 2014

April 16, 2014

Posted by **Nancy** on Thursday, August 2, 2012 at 1:18pm.

I was able to do this with the following technique:

1.1 x 6.022 x10^23 =6.62x10^23

1.4x10^10=7.358x10^15 minutes

I think then it is 6.62x10^23 x 7.358x10^15

But I want to learn how to do this problem using the formulas below instead of the .693/half life.

There are the following formulas:

starting amount) x (1/2)number of half-lives = ending amount (sometimes remaining rather than ending is used)

(1/2)number of half-lives is the decimal fraction which remains (1.000 is the original starting amount, 0.500 at the end of one half-life, 0.250 at the end of two, 0.125 at the end of three, etc.)

number of half-lives that have occurred can be expressed as (total time elasped ÷ length of half-life)

- chemistry -
**DrBob222**, Thursday, August 2, 2012 at 3:15pmI think what you want applies only to what remains or what decays and not to activity.

If we want to know what remains after 2 half lives it is (1/2^n) = 1/4 or after 3 half lives is (1/2^3) = 1/8 etc.

What you have here is activity at t=0.

R^{o}= kN^{o}

k = 0.693/t_{1/2}= 0.693/7.36E15 = about 9.4E-17

Ro = 9.4E-17 x 6.62E23 = ?dpm

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