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?

1 Answer
Johnson Z.
Mar 26, 2016

The sample is approximately #4065# years old.

Explanation:

When dealing with a half life question, it is best to use the half-life formula, which is expressed as:

#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.

  • In this case, #y=100%-39%=61%#, since #39%# of the sample decayed, leaving 61% to remain. Express #y# as #61#, derived from #61%#, but without the "#%#."
  • Express #a# in the equation as #100#, derived from #100%#, but without the "#%#."
  • The value of #b=100%-"decay rate"#, or in your case, #b=100%-50%=50%# or expressed as a fraction, #1/2#.

#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)|)))#

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