A bone was found to be approximately 24000 years old. The current amount of carbon-14 in the bone was 5.000 atoms. How many atoms must the bone have had originaly, assuming that the half-life of carbon-14 is approximately 6000 years?

2 Answers
Michael
Nov 22, 2016

About 80 atoms.

Explanation:

The equation for 1st order decay is:

#sf(N_t=N_0e^(-lambdat)#

Taking natural logs of both sides #rArr#

#sf(lnN_t=lnN_0-lamdat)#

#sf(lambda)# is the decay constant which is inversely related to the 1/2 life:

#sf(lambda=0.693/t_(1/2)=0.693/6000=0.0001155color(white)(x)a^(-1))#

We need to find #sf(N_0)# so:

#sf(ln(5)=ln(N_0)-0.0001155xx24000)#

#:.##sf(ln(N_0)=1.609+2.772=4.381)#

From which:

#sf(N_0=79.9)#

Oscar L.
Nov 23, 2016

Probably 80 thousand as opposed to 80.

Explanation:

Michael's answer is mathematically correct but (as I commented there), 5.000 may mean five thousand as rendered in Europe. Assuming the usual measurement techniques for C-14 we probably could not resolve five atoms in the measurenent but could resolve five thousand.

Related questions
See all questions in Nuclear Half-Life Calculations
Impact of this question
2760 views around the world
You can reuse this answer
Creative Commons License