1. Carbon-14(C14) decays at a constant rate in such a way that it reduces to 50% in 5568 years. Find the age of an old wooden piece in which the carbon is only 25% of the original.

Asked by rajubarman | 1st Dec, 2019, 10:03: AM

Expert Answer:

We know, 
Decay constant
λ= 0.693 /t1/2 
t1/2 = 5568 years ... (given)
Thus, 
λ= 0.693 /5568 = 1.24 × 10-4 / year
 
We know, 
By law of radioactive decay, 
N/N0 = e-λt 
The number of active nuclei at t = 0,  N0 = 100%
And N = 25% 
Thus, 
25/100 = e-λt 
or 
ln(0.25) = -λt
ln(0.25) = -1.24 × 10-4 × t
t = -1.3862/ (-1.24 × 10-4 ) ≈ 11180 years
 

Answered by Shiwani Sawant | 3rd Dec, 2019, 06:22: PM