Question

The half life of carbon 14 is 5730 years how old is a bone containing 20% of its original carbon 14

Answer 1

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

Answer 2

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