SME
Grade 12 Science · Earth & Space Science
Geologic time and radiometric dating
Use this for “A bone has 12.5% of its carbon-14 left. How old is it?” with C-14’s half-life of 5,730 years.
Diagram
After 17,200 years, 3 half-lives have passed: 100 × (1/2)³ = 12.5% left.
In the grid 13 of 100 C-14 atoms are left; each half-life, half of those still there decay.
This is an example. Type your own values to start a new problem.
p
%
=
100%
×
(0.5)
÷
n
Half-lives passed
worked out (not typed)
q
C-14 decayed
worked out (not typed) · %
Formulas
n
=
t
T
3
=
17,190
5,730
p
=
100
×
0.5
n
12.5
=
100
×
0.5
3
q
=
100
−
p
87.5 = 100 − 12.5
Assumptions
•
Living things keep the same C-14 level as the air; decay starts when they die.
•
After about 50,000 years too little is left to measure.
•
Only for once-living material: wood, bone, shell or cloth.
Step-by-step
WE KNOW
p = 12.5%, T = 5,730 years
FIND
half-lives passed (n), C-14 decayed (q), age (t)
Step 1 · Find half-lives passed (n)
Use p = 100 × 0.5ⁿ
Count
the
halvings:
n
is
the
power
of
2
that
equals
100
p
.
n
=
ln(
100
p
)/ln(2)
n
=
ln(100/12.5)/ln(2)
n
=
ln(8)/ln(2)
n
=
2.079/0.6931
n
=
3
Step 2 · Find C-14 decayed (q)
Use q = 100 − p
What is not left has decayed to N-14.
q
=
100
−
12.5
q
=
87.5%
Step 3 · Find age (t)
Use n = t ÷ T
Each
half-life
takes
T
:
multiply.
t
=
n
×
T
t
=
3
×
5,730
t
=
17,190
years
Check
Put the numbers back in. Both sides should match.
3
=
17,190
5,730
✓
12.5
=
100
×
0.5
3
✓
87.5 = 100 − 12.5 ✓
Topic check
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