Historical Geology · Problem type

Age from an isochron

Use this for “Samples of one granite plot on a line of slope 0.0145 on a graph of ⁸⁷Sr/⁸⁶Sr against ⁸⁷Rb/⁸⁶Sr. How old is the granite?”

Chart

y(x) = 0.0145x + 0.704501234560.720.740.760.780.8⁸⁷Rb/⁸⁶Sr⁸⁷Sr/⁸⁶Sr(0, 0.7045)(2, 0.7335)
y-intercept (0, 0.7045)
y(2) = 0.7335
This is an example. Type your own values to start a new problem.
t (Ma)
=
ln(1
+
m
)
÷
λ (yr⁻¹)
(⁸⁷Sr/⁸⁶Sr)₀
Initial ratio
example
x
A sample’s ⁸⁷Rb/⁸⁶Sr
example
y
That sample’s ⁸⁷Sr/⁸⁶Sr
calculated

Formulas

m
=
e
λ
×
t
−
1
0.0145
=
e
1.42 ×
10
-11
× (1.0138 ×
10
9
)
−
1
y
=
(⁸⁷Sr/⁸⁶Sr)₀
+
m
×
x
0.7335 = 0.7045 + 0.0145 × 2
Formulas are worked in yr. The step-by-step shows the conversions.

Assumptions

•
The samples formed together from one melt, so all began with the same ⁸⁷Sr/⁸⁶Sr: the initial ratio, where the line meets the axis.
•
Rb-87 decays to Sr-87 with λ = 1.42 × 10⁻¹¹ per year, so each point climbs by its ⁸⁷Rb/⁸⁶Sr times eλt − 1: the slope.
•
The slope grows with age alone, so no starting daughter needs guessing; the rock stayed closed since it formed.

Step-by-step

WE KNOW
m = 0.0145, λ = 1.42 × 10⁻¹¹ yr⁻¹, (⁸⁷Sr/⁸⁶Sr)₀ = 0.7045, x = 2
FIND
age (t), that sample’s ⁸⁷Sr/⁸⁶Sr (y)
Step 1 · Find age (t)
Use m = eλ × t − 1
Add 1 to the slope, take ln to undo the e, then divide by the decay constant.
t
=
ln(1 +
m
)
λ
t
=
ln(1 + 0.0145)
1.42 ×
10
-11
ln(1 + 0.0145) = 0.0144
t
=
1.0138 ×
10
9
yr
Step 2 · Find that sample’s ⁸⁷Sr/⁸⁶Sr (y)
Use y = (⁸⁷Sr/⁸⁶Sr)₀ + m × x
The sample started at the initial ratio and gained slope × ⁸⁷Rb/⁸⁶Sr of new Sr-87.
y
=
0.7045
+
0.0145
×
2
y
=
0.7045
+
0.029
y
=
0.7335
Step 3 · Convert the answers
Change each answer back to the unit you chose.
t = 1.0138 × 10⁹ yr = 1,014 Ma (1 Ma = 10⁶ yr)
Check
Put the numbers back in. Both sides should match.
0.0145
=
e
1.42 ×
10
-11
× (1.0138 ×
10
9
)
−
1
✓
0.7335 = 0.7045 + 0.0145 × 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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