Answer:
The radioactive isotope 14C has a half-life of approximately 5715 years. A piece of ancient charcoal contains only 75% as much of the radioac
Step-by-step explanation:
ece of modern charcoal. How long ago was the tree burned to make the ancient charcoal
The time required by the radioactive ancient charcoal to be consumed 25% has been 2,354.35 years.
The half-life can be defined as the time required by the substance to reduce to half of its initial concentration. The half-life can be expressed as:
Amount of substance remained = Initial concentration [tex]\rm \times\;\dfrac{1}{2}^\dfrac{t}{half-life}[/tex]
The t has been the time required.
For the ancient charcoal piece,
Let the initial concentration = 1
The remained charcoal = 75% = 0.75
The half-life of the 14C = 5715 years
Substituting the values:
0.75 = 1 [tex]\rm \times\;\dfrac{1}{2}^\dfrac{t}{5715}[/tex]
Taking log on both the side:
ln 0.75 = ln [tex]\rm \dfrac{1}{2}^\dfrac{t}{5715}[/tex]
ln 0.75 = [tex]\rm \dfrac{t}{5715}\;\times[/tex] ln 0.5
[tex]\rm \dfrac{ln\;0.75}{ln\;0.5}\;=\; \dfrac{t}{5715}[/tex]
[tex]\rm \dfrac{-0.124}{-0.301}\;=\;\dfrac{t}{5715}[/tex]
0.412 × 5715 = t
t = 2,354.35 years.
The time required by the radioactive ancient charcoal to be consumed 25% has been 2,354.35 years.
For more information about the half-life, refer to the link:
https://brainly.com/question/24710827
