Question

How long does it take for 25% of the C-14 atoms in a sample of C-14 to decay? If a sample of C-14 initially contains 1.5 millimol of C-14, how many millimoles are left after 2255 years?

The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration.

Answer 1

Answer 1: `t = 2378" yrs"`

Answer 2: `Q(2255" yrs") = 1.14" millimol"`

Explanation:

How long does it take for 25% of the C-14 atoms in a sample of C-14 to decay?

Starting with the equation:

`Q(t) = Q(0)(1/2)^(t/t_"half-life")" [1]"`

Digress for a while and use the formula for `Delta%`

`Delta% = 100(NewValue-OldValue)/(OldValue)`

Substitute `Delta% = -25%, NewValue = Q(t), and OldValue = Q(0)`

`-25% = 100(Q(t)-Q(0))/(Q(0))`

Divide both sides by 100:

`-0.25= (Q(t)-Q(0))/(Q(0))`

Separate into two fractions:

`-0.25= (Q(t))/(Q(0))-(Q(0))/(Q(0))`

The second fraction becomes -1:

`-0.25 = (Q(t))/(Q(0)) - 1`

Add 1 to both sides:

`(Q(t))/(Q(0)) = 0.75" [2]"`

Divide both sides of equation [1] by `Q(0)`:

`(Q(t))/(Q(0))=(1/2)^(t/t_"half-life")" [1.1]"`

Substitute equation [2] into equation [1.1]

`0.75=(1/2)^(t/t_"half-life")" [1.2]"`

Use the natural logarithm on both sides:

`ln(0.75)=ln((1/2)^(t/t_"half-life"))" [1.2]"`

Use the property of logarithms `ln(a^c) = (c)ln(a)`:

`ln(0.75)=(t/t_"half-life")ln(1/2)" [1.3]"`

`t = t_"half-life"ln(0.75)/ln(1/2)" [1.4]"`

Substitute `t_"half-life" = 5730" yrs"` into equation [1.4]:

`t = (5730" yrs")ln(0.75)/ln(1/2)" [1.5]"`

`t = 2378" yrs"`

If a sample of C-14 initially contains 1.5 millimol of C-14, how many millimoles are left after 2255 years?

`Q(2255" yrs") = (1.5xx10^-3" mol")(1/2)^((2255" yrs")/(5730" yrs")`

`Q(2255" yrs") = 1.14" millimol"`