Posted by **Angie** on Wednesday, April 15, 2009 at 4:28pm.

Cesium-137 has a half-life of 30 years. Suppose a lab stored a 30-mCi sample n 1973. How much will be left in 2063?

- Chemistry/math -
**DrBob222**, Wednesday, April 15, 2009 at 5:45pm
Use the half life to calculate the k in

k = 0.693/(t1/2).

Then use k in the following equation.

ln(Co/C) = kt

Co = 30 mCi

Solve for C. The answer will be in mCi.

k = from above

t = years from dates listed.

Post your work if you get stuck.

- Chemistry/math -
**Angie**, Wednesday, April 15, 2009 at 5:52pm
what does k stand for?

- Chemistry/math -
**DrBob222**, Wednesday, April 15, 2009 at 6:01pm
k is the proportionality constant for the equation:

ln(Co/C) = kt and it is evaluated from k = 0.693/t(1/2). In this case, its

k=0.693/30 years = ??

- Chemistry/math -
**Angie**, Wednesday, April 15, 2009 at 6:36pm
0.0231?

- Chemistry/math -
**DrBob222**, Wednesday, April 15, 2009 at 8:28pm
Yes, k = 0.0231 years^-1. Now put that into the ln formula I provided and solve for C following the instructions I gave you.

- Chemistry/math -
**Angie**, Wednesday, April 15, 2009 at 9:37pm
why is it to the -1?

- Chemistry/math -
**DrBob222**, Wednesday, April 15, 2009 at 11:34pm
Are you putting me on?

1/x = x^-1.

so 0.693/30 = 0.0231 years^-1. It's just a unit and the -1 means years is in the denominator. It's the same thing as saying wavenumber = 1/5 cm = 0.2 cm^-1 or in words it is 0.2 reciprocal centimeters.

- Chemistry/math -
**Angie**, Thursday, April 16, 2009 at 1:33pm
ok thanks

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