Question

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?

Answer 1

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)|)))`