Question

carbon-( c14) decays at a constant rate in such a way that introduced to 50% in 5568 years find the age of an old wooden piece in which the carbon is only 25% of the original .​

Answer 1

Answer:

11180 years

Step-by-step explanation:

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