Better Students Ask More Questions.
The half-life of carbon-14 is about 5370 years. What percent of the original carbon-14...
2 Answers | add yours
If t is the half life of a radioactive substance the percentage (p) of the radioactive material left after time T is given by:
p = 100*[0.5^(T/t)]
It is given:
Half life of carbon-14 = t = 5370 years, and
Actual time elapsed = T = 2500 years
Substituting these values in equation for p we get:
p = 100*[0.5^(2500/5370)] = 72.4195
72.4195 percent of carbon-14 will remain after 2500 years
Posted by krishna-agrawala on December 13, 2009 at 12:12 AM (Answer #1)
High School Teacher
The half life of carbon-14 = 5370 years.
Let m be the mass of the carbon-14, now
Therefore, the mass ofl carbon-14 to be left out after t tears = m(1/2)^ (t/5370), put t= 2500
=72.4195266% is the remaining carbon-14, after 2500 yrs
Posted by neela on December 12, 2009 at 4:29 PM (Answer #2)
Related QuestionsSee all »
Join to answer this question
Join a community of thousands of dedicated teachers and students.