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#### orthoformate

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« on: August 11, 2015, 12:49:18 PM »
A question on the GRE chemistry subject exam asks:

The half-life of 14C is 5730 years. The 14C activity of living material is approximately 920 decays/hr per gram of carbon. A fragment of wool fabric from an archaeological site has an activity of 680 decays/hr per gram of carbon. The approximate date of the sample is what?

a. 1950 A.D.
b. 500 B.C.
c. 3700 B.C.

I though that since the integrated rate equation is A=Ao*e^-kt then the dA/dt equation must be the derivative of this function

dA/dt=e^-kt

converting 680 decay/hr into years=

5,956,800 decays/years=e^-(0.00012)t

15.6=-0.00012t

t=130,000 years...this cant be right..

I feel like I must be going wrong early, does anyone have any advice?

#### Corribus

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« Reply #1 on: August 11, 2015, 02:50:11 PM »
No integration is required for this problem. Just use your rate equation and the values provided, but be sensitive to the fact that you'll have to adjust it for a different type of log scale. (I.e., half-life is not a natural logarithmic decay).

What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### billnotgatez

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« Reply #2 on: August 11, 2015, 03:28:18 PM »
When taking tests I sometimes do a quick reality check.
If I started at 920 decays/hr per gram of carbon what would be the decays/hr per gram of carbon after 5730 years?
after 11460 years?
Then I would have a  relative check on the answer I got through other calculations.
« Last Edit: August 11, 2015, 03:39:15 PM by billnotgatez »

#### orthoformate

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« Reply #3 on: August 12, 2015, 11:36:53 AM »
Corribus,

Do you mean use this eqn? dA/dt= -kA

Or this one: A=Aoe-kt

why is it necessary to adjust to base 10 log?

billnotgatez,

Do you mean that after 5730 years, the rate should also be 460 decays/hr per gram?

Thanks for the responses!

#### Corribus

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« Reply #4 on: August 12, 2015, 11:48:50 AM »
Or this one: A=Aoe-kt
This one.

Quote
why is it necessary to adjust to base 10 log?

The natural log scale specifies the amount of time it takes for the concentration to drop from the initial concentration to a concentration of 1/e, not 1/2. This is the difference between half-life and the quantity τ, the reciprocal of the rate constant, k. t1/2 is equal to τ ln (2), or ln (2) / k. So, you solve the problem for the rate constant in the normal way, and then you just have to take this into account when you do your conversion to half-life.

Quote
Do you mean that after 5730 years, the rate should also be 460 decays/hr per gram?
This is what he means. With this type of estimation, you should be able to pick the right answer without really calculating anything at all.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman