Q1: A living plant contains approximately the same isotopic abundance of C-14 as does atmospheric carbon dioxide. The observed rate of decay of C-14 from a living plant is 15.3 disintegrations per minute per gram of carbon. How much disintegration per minute per gram of carbon will be measured from a 12900-year-old sample? (The half-life of C-14 is 5730 years.)
Given:
Initial activity, A₀ = 15.3 disintegrations/min/g
Time, t = 12900 years
Half-life, T₁/₂ = 5730 years
Formula: A = A₀ (1/2)^(t / T₁/₂)
Calculation:
Number of half-lives, n = t / T₁/₂ = 12900 / 5730 = 2.25
A = 15.3 × (1/2)^(2.25)
(1/2)^(2.25) = 2^(-2.25) = 1 / 2^(2.25)
2^(2.25) = 2² × 2^(0.25) = 4 × 1.1892 = 4.7568
(1/2)^(2.25) = 1 / 4.7568 = 0.2102
A = 15.3 × 0.2102
A = 3.216 disintegrations/min/g
Answer: 3.22 disintegrations/min/g
Initial activity, A₀ = 15.3 disintegrations/min/g
Time, t = 12900 years
Half-life, T₁/₂ = 5730 years
Formula: A = A₀ (1/2)^(t / T₁/₂)
Calculation:
Number of half-lives, n = t / T₁/₂ = 12900 / 5730 = 2.25
A = 15.3 × (1/2)^(2.25)
(1/2)^(2.25) = 2^(-2.25) = 1 / 2^(2.25)
2^(2.25) = 2² × 2^(0.25) = 4 × 1.1892 = 4.7568
(1/2)^(2.25) = 1 / 4.7568 = 0.2102
A = 15.3 × 0.2102
A = 3.216 disintegrations/min/g
Answer: 3.22 disintegrations/min/g
Q2: The smallest C-14 activity that can be measured is about 0.20%. If C-14 is used to date an object, the object must have died within how many years?
Given:
Activity remaining, A/A₀ = 0.20% = 0.0020
Half-life of C-14, T₁/₂ = 5730 years
Formula: A/A₀ = (1/2)^(t / T₁/₂)
Taking log on both sides:
ln(A/A₀) = (t / T₁/₂) × ln(1/2)
t = T₁/₂ × [ln(A/A₀) / ln(1/2)]
Calculation:
ln(0.0020) = ln(2 × 10⁻³) = ln2 + ln(10⁻³)
ln(0.0020) = 0.693 - (3 × 2.3026)
ln(0.0020) = 0.693 - 6.9078
ln(0.0020) = -6.2148
ln(1/2) = -0.693
t = 5730 × [(-6.2148) / (-0.693)]
t = 5730 × 8.967
t = 51380 years
Answer: 51380 years
Activity remaining, A/A₀ = 0.20% = 0.0020
Half-life of C-14, T₁/₂ = 5730 years
Formula: A/A₀ = (1/2)^(t / T₁/₂)
Taking log on both sides:
ln(A/A₀) = (t / T₁/₂) × ln(1/2)
t = T₁/₂ × [ln(A/A₀) / ln(1/2)]
Calculation:
ln(0.0020) = ln(2 × 10⁻³) = ln2 + ln(10⁻³)
ln(0.0020) = 0.693 - (3 × 2.3026)
ln(0.0020) = 0.693 - 6.9078
ln(0.0020) = -6.2148
ln(1/2) = -0.693
t = 5730 × [(-6.2148) / (-0.693)]
t = 5730 × 8.967
t = 51380 years
Answer: 51380 years
Q3: How long will it take for 25% of the C-14 atoms in a sample of C-14 to decay?
Given:
25% decayed means 75% remaining
A/A₀ = 0.75
Half-life of C-14, T₁/₂ = 5730 years
Formula: A/A₀ = (1/2)^(t / T₁/₂)
Taking log on both sides:
ln(A/A₀) = (t / T₁/₂) × ln(1/2)
t = T₁/₂ × [ln(A/A₀) / ln(1/2)]
Calculation:
ln(0.75) = -0.2877
ln(1/2) = -0.693
t = 5730 × [(-0.2877) / (-0.693)]
t = 5730 × 0.4152
t = 2378 years
Answer: 2378 years
25% decayed means 75% remaining
A/A₀ = 0.75
Half-life of C-14, T₁/₂ = 5730 years
Formula: A/A₀ = (1/2)^(t / T₁/₂)
Taking log on both sides:
ln(A/A₀) = (t / T₁/₂) × ln(1/2)
t = T₁/₂ × [ln(A/A₀) / ln(1/2)]
Calculation:
ln(0.75) = -0.2877
ln(1/2) = -0.693
t = 5730 × [(-0.2877) / (-0.693)]
t = 5730 × 0.4152
t = 2378 years
Answer: 2378 years
Q4: The carbon-14 decay rate of a sample obtained from a young tree is 0.296 disintegration per second per gram of the sample. Another wood sample prepared from an object recovered at an archaeological excavation gives a decay rate of 0.109 disintegration per second per gram of the sample. What is the age of the object?
Given:
Initial activity, A₀ = 0.296 disintegrations/s/g
Current activity, A = 0.109 disintegrations/s/g
Half-life of C-14, T₁/₂ = 5730 years
Formula: A = A₀ e^(-ฮปt)
ฮป = ln2 / T₁/₂
t = (1/ฮป) × ln(A₀/A)
Calculation:
ฮป = 0.693 / 5730 = 1.209 × 10⁻⁴ year⁻¹
A₀/A = 0.296 / 0.109 = 2.716
ln(2.716) = 1.0
t = (1 / 1.209 × 10⁻⁴) × 1.0
t = 8271 years
Answer: 8271 years
Initial activity, A₀ = 0.296 disintegrations/s/g
Current activity, A = 0.109 disintegrations/s/g
Half-life of C-14, T₁/₂ = 5730 years
Formula: A = A₀ e^(-ฮปt)
ฮป = ln2 / T₁/₂
t = (1/ฮป) × ln(A₀/A)
Calculation:
ฮป = 0.693 / 5730 = 1.209 × 10⁻⁴ year⁻¹
A₀/A = 0.296 / 0.109 = 2.716
ln(2.716) = 1.0
t = (1 / 1.209 × 10⁻⁴) × 1.0
t = 8271 years
Answer: 8271 years
