The half-life of carbon-14 is 5700 years. What is the age to the nearest year of a sample in which 39% of the radioactive nuclei originally present have decayed?

Question

Naomi Aronson · Accepted Answer

The sample is approximately #4065# years old.

It is recommended to apply the half-life formula, which can be written as follows, when answering questions about half lives: #color(blue)(|bar(ul(color(white)(a/a)y=a(b)^(t/h)color(white)(a/a)|)))# where: #y=#final amount #a=#inital amount #b=#growth/decay #t=#time elapsed #h=#half-life #1#. Start by substituting your known values into the formula. #y=a(b)^(t/h)# #61=100(1/2)^(t/5700)# #2#. Divide both sides by #100#. #61/100=(1/2)^(t/5700)# #3#. Since the bases on both sides of the equation are not the same, take the logarithm of both sides. #log(61/100)=log((1/2)^(t/5700))# #4#. Use the log property, #log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)#, to rewrite the right side of the equation. #log(61/100)=t/5700*log(1/2)# #5#. Isolate for #t#. #t/5700=(log(61/100))/(log(1/2))# #t=(5700*log(61/100))/(log(1/2))# #6#. Solve for #t#. #t=4064.78...# #color(green)(|bar(ul(color(white)(a/a)t~~4065color(white)(i)"years old"color(white)(a/a)|)))#

Cooper Anderson · Answer

undefined To find the age of the sample, you can use the formula for radioactive decay:

$$ N_t = N_0 \times (1/2)^{t/T} $$

Where:
- $ N_t $ = the amount of substance remaining after time t
- $ N_0 $ = the initial amount of substance
- $ t $ = time elapsed
- $ T $ = half-life

Rearrange the formula to solve for $ t $:

$$ t = \frac{T \times \log_2(N_t/N_0)}{\log_2(1/2)} $$

Given:
- $ T = 5700 $ years
- $ N_t/N_0 = 0.39 $

Substitute the values and solve for $ t $.