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A sample contained 1.00 g of 14C exactly 2,000 years ago. (1)

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A sample contained 1.00 g of 14C exactly 2,000 years ago.
(1) How much 14C is present today (g)?
(2) What is the activity of the 14C present in the sample today (use units of Bq)? You will need to know that the half-life of 14C is 5,730 years, and obtain a value for the decay constant k. This constant has units of reciprocal time (years-1) and the value for k can be obtained by using the following first order decay relationship:
Cx = C0e-kt Equation (1)
Here Cx is the concentration of anything that decays (or is eliminated in some other way) at a constant rate that is proportional to the amount of the substance present (e.g. 14C), and C0 is the amount of the same material originally present. The rate constant ¬k has units of reciprocal time as already stated, and the variable t is elapsed time (0 to time x). Since after 5,730 years exactly ½ of the original 14C is gone we can set up the equation above to determine k:
C5,730 = 1/2C0 Equation (2)
No matter how much 14C was originally present, there will be exactly one-half the amount left after one half-life period (5,730 years), so we can write this equation as follows and solve for k:
Thus, 0.5 = C0e(-k/years) (5,730 years) Equation (3)
Now use the value of k you just obtained by using natural logarithms and algebra, and Equation (1) to answer the first question. Look up the activity of 14C to solve the second question.
Kindly indicate dead line for replying this question. Do you have some information on activity?
(1)N=N0exp(-kt)

N=N0/2, t = 5730

Hence, 1/2 exp(-5730k) or k = ln/5730

Hence, N =1*exp(-ln2*2000/5730) = 0.7851

Hence, 0.7851 gm will remain after 2000 years.

(2) Activity =|dN/dt|=kN=(ln2/5730)*2000*365*24*3600 Bq=7.63*10^6 Bq Ans.

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Customer: replied 4 years ago.
Thanks for your help.
Most welcome.
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