SME
Grade 12 Science · Problem type
Dating volcanic ash with potassium-argon
Use this for “Volcanic ash holds 0.01 argon-40 atom for every potassium-40 atom. How old is the ash?”
Diagram
91.5% of the potassium-40 is left, drawn as 91 of 100 atoms.
Of the 9 that decayed, 10.7% became argon‑40 (1) and 89.3% calcium‑40 (8).
This is an example. Type your own values to start a new problem.
P
%
=
100%
÷
(1
+
R
÷
0.107)
=
100%
×
(0.5)
÷
1,250
Only 10.7% of potassium-40 decays make argon-40, so R ÷ 0.107 atoms decayed for each one left: P = 100% ÷ (1 + R ÷ 0.107). The half-lives in P then give the age.
n
Half-lives passed
worked out (not typed)
Formulas
P
=
100
1 +
R
÷ 0.107
91.5
=
100
1 + 0.01 ÷ 0.107
P
=
100
×
0.5
n
91.5
=
100
×
0.5
0.129
t
=
n
×
1,250
161 = 0.129 × 1,250
Assumptions
•
Potassium-40 has a half-life of 1,250 million years: 10.7% of its decays make argon-40 and 89.3% make calcium-40.
•
Argon is a gas that escapes molten rock but is trapped once the ash cools, so the clock starts at the eruption.
•
Only the argon is counted: rock is already full of calcium-40, so the new calcium cannot be told apart.
•
Nothing on Earth is older than about 4,570 million years, so R is at most about 1.24.
Step-by-step
WE KNOW
R = 0.01
FIND
potassium-40 left (P), half-lives passed (n), age of the ash (t)
Step 1 · Find potassium-40 left (P)
Use P = 100 ÷ (1 + R ÷ 0.107)
Each
argon
atom
stands
for
1
0.107
decayed
potassium
atoms,
so
R
0.107
atoms
decayed
for
each
one
left.
P
=
100
1 + 0.01 ÷ 0.107
P
=
100
1 + 0.093458
P
=
100
1.0935
P
=
91.45%
Step 2 · 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(100/91.45)/ln(2)
n
=
ln(1.0935)/ln(2)
n
=
0.089384/0.69315
n
=
0.129
Step 3 · Find age of the ash (t)
Use t = n × 1,250
The age is the half-lives passed times the half-life.
t
=
0.129
×
1,250
t
=
161
million
years
Check
Put the numbers back in. Both sides should match.
91.45
≈
100
1 + 0.01 ÷ 0.107
✓
91.45
≈
100
×
0.5
0.129
✓
161 ≈ 0.129 × 1,250 ✓
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