Class 10 - Unit # 20 : Nuclear Structure - Solved Numericals

Radioactive Decay & Carbon Dating

1. 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 $15.3\text{ disintegrations per minute per gram}$ of carbon. How many disintegrations per minute per gram of carbon will be measured from a $12900\text{-year-old}$ sample? (The half-life of C-14 is $5730\text{ years}$.)

Data:

• Initial decay rate ($A_0$) = $15.3\text{ dpm/g}$
• Age of the sample ($t$) = $12900\text{ years}$
• Half-life of C-14 ($T_{1/2}$) = $5730\text{ years}$
• Remaining decay rate ($A$) = ?

Solution:

First, we calculate the number of half-lives ($n$) that have elapsed during the given time:

$$n = \frac{t}{T_{1/2}} = \frac{12900}{5730} \approx 2.2513$$

Now, the fraction of activity remaining after $n$ half-lives is determined by:

$$\text{Fraction Remaining} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^{2.2513} \approx 0.210$$

Using the fraction to find the final active decay rate ($A$):

$$A = A_0 \cdot \left(\frac{1}{2}\right)^n$$ $$A = 15.3 \times 0.210 \approx 3.21\text{ dpm/g}$$

RESULT: The fraction remaining after this time is $0.21$ and the final measured activity will be $3.21\text{ disintegrations per minute per gram}$.

2. The smallest C-14 activity that can be measured is about $0.20\%$ of its initial value. If C-14 is used to date an object, the object must have died within how many years?

Data:

• Remaining percentage activity = $0.20\%$
• Remaining fraction ($\frac{A}{A_0}$) = $\frac{0.20}{100} = 0.0020$
• Half-life of C-14 ($T_{1/2}$) = $5730\text{ years}$
• Maximum measurable age ($t$) = ?

Solution:

We use the standard decay fraction formula:

$$\frac{A}{A_0} = \left(\frac{1}{2}\right)^n \implies 0.0020 = (0.5)^n$$

Taking the common logarithm ($\log_{10}$) on both sides:

$$\log_{10}(0.0020) = \log_{10}\left((0.5)^n\right)$$ $$\log_{10}(0.0020) = n \cdot \log_{10}(0.5)$$ $$-2.6989 = n \cdot (-0.3010)$$ $$n = \frac{-2.6989}{-0.3010} \approx 8.9664$$

Now, calculating the time period ($t$) corresponding to these half-lives:

$$t = n \cdot T_{1/2}$$ $$t = 8.9664 \times 5730 \approx 51378\text{ years}$$

RESULT: The object must have died within $51378\text{ years}$ (approx $51375\text{ years}$) to be dateable.

3. How long will it take for $25\%$ of the C-14 atoms in a sample of C-14 to decay?

Data:

• Percentage of sample decayed = $25\%$
• Remaining percentage activity = $100\% - 25\% = 75\%$
• Remaining fraction ($\frac{A}{A_0}$) = $\frac{75}{100} = 0.75$
• Half-life of C-14 ($T_{1/2}$) = $5730\text{ years}$
• Time required ($t$) = ?

Solution:

Setting up the decay fraction equation:

$$\frac{A}{A_0} = \left(\frac{1}{2}\right)^n \implies 0.75 = (0.5)^n$$

Taking the common logarithm ($\log_{10}$) on both sides:

$$\log_{10}(0.75) = \log_{10}\left((0.5)^n\right)$$ $$\log_{10}(0.75) = n \cdot \log_{10}(0.5)$$ $$-0.1249 = n \cdot (-0.3010)$$ $$n = \frac{-0.1249}{-0.3010} \approx 0.4150$$

Now, computing the elapsed time ($t$):

$$t = n \cdot T_{1/2}$$ $$t = 0.4150 \times 5730 \approx 2378\text{ years}$$

RESULT: The time required for $25\%$ of the sample to decay is $2378\text{ years}$.

4. The carbon-14 decay rate of a sample obtained from a young tree is $0.296\text{ disintegrations per second per gram}$ of the sample. Another wood sample prepared from an object recovered at an archaeological excavation gives a decay rate of $0.109\text{ disintegrations per second per gram}$ of the sample. What is the age of the object?

Data:

• Initial decay rate ($A_0$) = $0.296\text{ dps/g}$
• Present decay rate ($A$) = $0.109\text{ dps/g}$
• Half-life of C-14 ($T_{1/2}$) = $5730\text{ years}$
• Age of the archaeological object ($t$) = ?

Solution:

Calculating the ratio of remaining activity:

$$\frac{A}{A_0} = \frac{0.109}{0.296} \approx 0.3682$$

Applying the exponential decay relation $\frac{A}{A_0} = (0.5)^n$:

$$0.3682 = (0.5)^n$$ $$\log_{10}(0.3682) = n \cdot \log_{10}(0.5)$$ $$-0.4339 = n \cdot (-0.3010)$$ $$n = \frac{-0.4339}{-0.3010} \approx 1.4415$$

Calculating the actual age ($t$):

$$t = n \cdot T_{1/2}$$ $$t = 1.4415 \times 5730 \approx 8260\text{ years}$$

RESULT: The age of the recovered archaeological object is approximately $8260\text{ years}$ (calculated as $8258\text{ years}$ dynamically).