Question

When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of 14C per gram per minute. A sample from an ancient piece of charcoal shows 14C activity to be 12.3 disintegrations per gram per minute. How old is this sample? Half-life of 14C is 5730 y.

Answer 1

Explanation:

It is given:

Charcoal has the initial activity, which is denoted as $A_{0}=15.3$disintegrations per minute per gram.

Charcoal has the half-life, $T 12=5730$ years

After a few years, the charcoal’s final activity, A = 12.3 disintegrations per minute per gram

Constant of disintegration,

$\lambda=0.693 \mathrm{T} 12=0.6935370 \mathrm{y}-1$

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,

$A=A O e-\lambda t$

$A=A O e-0.6935730 \times t$

$\Rightarrow \mathrm{t}=1804.3 \text { years }$

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