SME
Historical Geology · Problem type
Potassium–argon age with the branching decay
Use this for “A lava flow holds 0.01 argon-40 atom for every potassium-40 atom. How long ago did it cool?”
Diagram
91.3% of the potassium-40 is left, drawn as 91 of 100 atoms.
Of the 9 that decayed, 10.5% became argon‑40 (1) and 89.5% calcium‑40 (8).
This is an example. Type your own values to start a new problem.
t (Ma)
=
(1
÷
λ)
ln(1
+
Ar/K
÷
0.105)
p
Potassium-40 left
calculated · %
Formulas
t
=
(
1
0.000554
)
×
ln(1
+
Ar/K
0.105
)
164
=
(
1
0.000554
)
×
ln(1
+
0.01
0.105
)
p
=
100
1 + (Ar/K) ÷ 0.105
91.3
=
100
1 + 0.01 ÷ 0.105
Assumptions
•
K-40 decays with λ = 5.54 × 10⁻¹⁰ per year (0.000554 per million years); 10.5% of its decays give argon-40, the rest calcium-40.
•
Argon is a gas, so heating drives it out and resets the clock: the age is when the rock last cooled.
•
No argon was trapped as the rock cooled and none has leaked out since.
•
Earth is about 4,600 million years old, so no rock here is older.
Step-by-step
WE KNOW
Ar/K = 0.01
FIND
age (t), potassium-40 left (p)
Step 1 · Find age (t)
Use t = (1 ÷ 0.000554) × ln(1 + (Ar/K) ÷ 0.105)
Each
argon
atom
stands
for
1
0.105
decayed
K-40
atoms.
Add
1
for
the
atom
left,
take
ln,
then
divide
by
λ.
t
=
(
1
0.000554
)
×
ln(1
+
0.01
0.105
)
0.01
0.105
=
0.09524
ln(1 + 0.09524) = 0.09097
t
=
164
Ma
Step 2 · Find potassium-40 left (p)
Use p = 100 ÷ (1 + (Ar/K) ÷ 0.105)
Ar/K ÷ 0.105 atoms decayed for each K-40 atom left, so the K-40 is 1 of every 1 + Ar/K ÷ 0.105 atoms.
p
=
100
1 + 0.01 ÷ 0.105
p
=
100
1 + 0.095238
p
=
100
1.0952
p
=
91.3%
Check
Put the numbers back in. Both sides should match.
164
=
(
1
0.000554
)
×
ln(1
+
0.01
0.105
)
✓
91.3
=
100
1 + 0.01 ÷ 0.105
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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