please solve this question---
carbon-14 (c14) decays at a constant rate in such a way that it reduces to 50% in 5,568 years . find the age of an old wooden piece in which the carbon is only 25% of the original?
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To find the time, we need to have the initial and final amount of the substance and t1/2 (all known) and value of rate constant k. k is calculated from t1/2 value.
Step 1: Determine k using formula k= 0.693/t1/2 t1/2 = 5568 years
K = 0.693/5568 =1.24 X 10-4 year-1
Step2: Find the time t, using equation N/N0 = -ekt
N0= 100% N = 12.5% t1/2 = 5568 years
12.5/100 = -ekt
ln 0.125 = -k X t
-2.07944= - k X t
t= 2.0794/1.24 X 10-4 = 16707.54 years
An easy way to calculate is to find how many half-lives are involved.
If we start with 100%, first half-life will reduce the amount to 50% of the initial amount.
Second and third half-life will reduce the amount to 25% and 12.5%.
So, 3 half-lives will reduce amount to 12.5%
Time taken for 3 half-lives = 3 X 5568 = 16704 years
Step 1: Determine k using formula k= 0.693/t1/2 t1/2 = 5568 years
K = 0.693/5568 =1.24 X 10-4 year-1
Step2: Find the time t, using equation N/N0 = -ekt
N0= 100% N = 12.5% t1/2 = 5568 years
12.5/100 = -ekt
ln 0.125 = -k X t
-2.07944= - k X t
t= 2.0794/1.24 X 10-4 = 16707.54 years
An easy way to calculate is to find how many half-lives are involved.
If we start with 100%, first half-life will reduce the amount to 50% of the initial amount.
Second and third half-life will reduce the amount to 25% and 12.5%.
So, 3 half-lives will reduce amount to 12.5%
Time taken for 3 half-lives = 3 X 5568 = 16704 years
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