given that carbon 14( C 14) decays at a constant rate in such a way that it reduces to 50 % in 5568years. find the age of an old wooden peice in which the carbon is only 12 1/2 % of the original
Answers
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
