SME
Historical Geology · Problem type
Radiocarbon age from activity
Use this for “Charcoal from a hearth gives 3.4 decays per minute per gram, and living carbon gives 13.6. How old is the charcoal?”
Diagram
After 11,500 yr, 2 half-lives have passed: 13.6 × (1/2)² = 3.4 dpm/g left.
In the grid 25 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.
t (yr)
=
5730
×
ln(1
÷
F
)
÷
ln
2
A
Activity
example · dpm/g
A₀
Living activity
example · dpm/g
Formulas
F
=
A
A₀
0.25
=
3.4
13.6
t
=
5,730
×
ln(1 ÷
F
)
ln(2)
11,460
=
5,730
×
ln(1 ÷ 0.25)
ln(2)
Assumptions
•
Only for once-living carbon (wood, bone, shell) younger than about 50,000 years; older samples have too little C-14 left to count.
•
C-14 decays to N-14 with a half-life of 5,730 years; living carbon gives about 13.6 decays per minute per gram (dpm/g).
•
The air’s C-14 is taken as steady over time; labs calibrate the age against tree rings for that.
Step-by-step
WE KNOW
A = 3.4 dpm/g, A₀ = 13.6 dpm/g
FIND
fraction of C-14 left (F), age (t)
Step 1 · Find fraction of C-14 left (F)
Use F = A ÷ A₀
Activity counts the C-14 atoms, so the share left is today’s activity over living carbon’s.
F
=
3.4
13.6
F
=
0.25
Step 2 · Find age (t)
Use t = 5,730 × ln(1 ÷ F) ÷ ln(2)
Each
half-life
halves
the
C-14,
so
ln(1 ÷
F
)
ln 2
counts
the
half-lives;
times
5,730
years
each.
t
=
5,730
×
ln(1 ÷ 0.25)
ln(2)
1
0.25
=
4
ln(4)
ln(2)
=
2
half-lives
t
=
11,460
yr
Check
Put the numbers back in. Both sides should match.
0.25
=
3.4
13.6
✓
11,460
=
5,730
×
ln(1 ÷ 0.25)
ln(2)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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