When a living organism dies, the amount of Carbon-14 in its system begins to decrease exponentially. Given that the half-life of Carbon-14 is 5,730 years, answer the following: a.) What is the exact value of the rate of decay for Carbon-14?

Question

b.) If a bone is found to contain 61% of the Carbon-14 in an equivalent living organism, how old is the bone? (Round to the nearest year)
The bone is approximately years old.

c.) After 6250 years what percent of Carbon-14 will remain in the bone? (Round to the nearest tenth of a percent)
Approximately
% will remain after 6250 years.

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Answer 1 · 2018-01-14T23:18:48

#y(t) = a * (1/2)^(t/5730)#
#4,086# years
#47.0%#

#y(t) = a * (1/2)^(kt)#
where #t# is in years

#y(5730) = a * (1/2)^(1) = 1/2a#

#y(5730) = a * (1/2)^((1/5730)*5730)#

this means that #k# is #1/5730#.

#y(t) = a * (1/2)^((1/5730)*t)#

#y(t) = a * (1/2)^(t/5730)#

-

#y(t) = a * (1/2)^(t/5730)#
where #t# is in years

#y(t) = 0.61a#

#a * (1/2)^(t/5730) = 0.61a#

#(1/2)^(t/5730) = 0.61#

#t/5730 = log_(1/2) 0.61#

#t = log_(1/2) 0.61 * 5730 #

#t = 4,086# (nearest #1#)

#y(t) = a * (1/2)^(t/5730)#

#y(6250) = a * (1/2)^(6250/5730)#

#(1/2)^(6250/5730) = 0.469517...#

#0.469517 * 100% = 47.0%# (nearest #0.1%#)