The half life of carbon 14 is 5730 years how old is a bone containing 20% of its original carbon 14
Answers
Answer:
13305 years
Explanation:
0.20X=X%281%2F2%29%5E%28T%2F5730%29
0.20=%281%2F2%29%5E%28T%2F5730%29
log%281%2F2%2C%280.2%29%29=T%2F5730
T%2F5730=%28log%28%280.2%29%29%2Flog%28%280.5%29%29%29
T=5730%2A%28log%28%280.2%29%29%2Flog%28%280.5%29%29%29
T=5730%2A2.322
T=13305years
Explanation:
It is given:
Charcoal has the initial activity, which is denoted as A_{0}=15.3A
0
=15.3 disintegrations per minute per gram.
Charcoal has the half-life, T 12=5730T12=5730 years
After a few years, the charcoal’s final activity, A = 12.3 disintegrations per minute per gram
Constant of disintegration,
\lambda=0.693 \mathrm{T} 12=0.6935370 \mathrm{y}-1λ=0.693T12=0.6935370y−1
For the action to attain 12.3 disintegrations per minute per gram, let the time taken by the sample at a time of t year.
Sample’s activity,
A=A O e-\lambda tA=AOe−λt
A=A O e-0.6935730 \times tA=AOe−0.6935730×t
\Rightarrow \mathrm{t}=1804.3 \text { years }⇒t=1804.3 years