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.

1 Answer
Oct 26, 2017

Answer 1: t = 2378" yrs"t=2378 yrs

Answer 2: Q(2255" yrs") = 1.14" millimol"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]"Q(t)=Q(0)(12)tthalf-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"