The radioactive isotope, _6^14C does not occur naturally but it is found at cons

Question

The radioactive isotope, _6^14C does not occur naturally but it is found at constant rate by the action of rays on the atmosphere. It is tanken up by plants and animals and deposited in the body structure along with natural carbon, but this process stops at death. The charcoal from the fire pit of an ancient camp has an activity due to _6^14C of 12.9 disintegrations per minute, per gram of carbon. If the percentage of _6^14C compared with normal Carbon in living trees is 1.35 the decay 10^-10^, the decay constant is 3.92 times 10^-10 s^-1 and the atomic weight = 12.0, what is the age of the campsite?

Explanation / Answer

given,activity of fire pit charcoal, a = 12.9 disintegrations per minute per gram of carbon
percentage of 14C compared to 12C in living trees, f = 1.35*10^-10 percent
decay constant of 14C , lambda = 3.92*10^-10 per s

consider 1 gram of carbon from the tree at t = 0
numer of carbon atoms = 1/12 mole = 6.602*10^23/12 = 0.551*10^23
amount of 14C in this , No= 1.35*10^-10*0.551*10^23*0.01 = 7.4475*10^10 atoms
now we know that
at any time t
N = No*e^(-lambda*t)
and activity at any point
A = No*lambda*e^(lambda*t) = N*lambda

A = 12.9 /60 disintegrations per second
12.9/60 = 7.4475*10^10 * e^(-3.92*10^-10 t)
t = 6.7782*10^10 s = 2147.906 years
so the campsite is 2147.906 years old

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.