The smallest 14C/12C ratio that can be reliably measured is about 3.0 * 10-15, setting a limit on the oldest carbon specimens that can be dated. How old would a sample with this carbon ratio be?
The correct answer and explanation is:
Let’s solve this step by step.
Given:
- Smallest measurable 14C12C\frac{^{14}C}{^{12}C} ratio = 3.0×10−153.0 \times 10^{-15}
- We want to find the age of a sample with this ratio.
Background:
Radiocarbon dating relies on the decay of 14C{}^{14}C (radioactive carbon isotope) in organic samples. Living organisms maintain a certain ratio of 14C{}^{14}C to 12C{}^{12}C in their tissues, roughly constant with the atmosphere.
Once the organism dies, no new 14C{}^{14}C is absorbed, and the 14C{}^{14}C decays exponentially over time: N(t)=N0e−λtN(t) = N_0 e^{-\lambda t}
Where:
- N(t)N(t) = 14C{}^{14}C atoms at time tt,
- N0N_0 = initial 14C{}^{14}C atoms (ratio when organism died),
- λ\lambda = decay constant of 14C{}^{14}C,
- tt = time elapsed since death.
Step 1: Initial 14C/12C{}^{14}C/{}^{12}C ratio
The modern ratio (at time of death) is approximately: (14C12C)0=1.3×10−12\left(\frac{{}^{14}C}{{}^{12}C}\right)_0 = 1.3 \times 10^{-12}
This is a standard accepted value from literature.
Step 2: Decay constant λ\lambda
The half-life of 14C{}^{14}C is approximately: t1/2=5730 yearst_{1/2} = 5730 \text{ years}
Decay constant λ\lambda is: λ=ln2t1/2=0.6935730=1.2097×10−4 year−1\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{5730} = 1.2097 \times 10^{-4} \text{ year}^{-1}
Step 3: Use the decay formula for the ratio:
(14C12C)t=(14C12C)0e−λt\left(\frac{{}^{14}C}{{}^{12}C}\right)_t = \left(\frac{{}^{14}C}{{}^{12}C}\right)_0 e^{-\lambda t}
Given: 3.0×10−15=1.3×10−12×e−λt3.0 \times 10^{-15} = 1.3 \times 10^{-12} \times e^{-\lambda t}
Solve for tt: e−λt=3.0×10−151.3×10−12=0.0023077e^{-\lambda t} = \frac{3.0 \times 10^{-15}}{1.3 \times 10^{-12}} = 0.0023077
Taking natural logarithm on both sides: −λt=ln(0.0023077)=−6.070-\lambda t = \ln(0.0023077) = -6.070 t=6.070λ=6.0701.2097×10−4=50,203 yearst = \frac{6.070}{\lambda} = \frac{6.070}{1.2097 \times 10^{-4}} = 50,203 \text{ years}
Answer:
A sample with a 14C/12C{}^{14}C/{}^{12}C ratio of 3.0×10−153.0 \times 10^{-15} would be approximately 50,200 years old.
Explanation:
Radiocarbon dating depends on measuring the 14C{}^{14}C isotope, which decays over time. The initial ratio in living things is about 1.3×10−121.3 \times 10^{-12}. Because 14C{}^{14}C decays exponentially with a half-life of 5730 years, the ratio decreases steadily.
When the 14C/12C{}^{14}C/{}^{12}C ratio falls below 3.0×10−153.0 \times 10^{-15}, it’s difficult to measure accurately due to background noise and instrument limits. This sets a practical dating limit for radiocarbon dating — about 50,000 years. Beyond this age, the amount of 14C{}^{14}C left is too small to detect reliably, so other dating methods must be used.
This age limit is why radiocarbon dating is best for dating relatively recent ancient samples (up to ~50,000 years). Samples older than this require other techniques like potassium-argon or uranium-series dating.