The `beta-` activity of a sample of `CO_(2)` prepared form a contemporary wood gave a count rate of `25.5` counts per minute `(cp m)`. The same of `CO_(2)` form an ancient wooden statue gave a count rate of `20.5 cp m`, in the same counter condition. Calculate its age to the nearest 50 year taking `t_(1//2)` for `.^(14)C` as 5770 year. What would be the expected count rate of an identical mass of `CO_(2)` form a sample which is 4000 year old?

Answer

To solve the problem, we need to calculate the age of the ancient wooden statue based on the count rates of the carbon-14 activity in two samples of CO₂. We will also determine the expected count rate for a CO₂ sample that is 4000 years old.

### Step 1: Understand the relationship between count rates and age
The count rate (R) of a radioactive sample is directly proportional to the number of radioactive nuclei present (N). The relationship can be expressed as:
\[ \frac{R_0}{R} = \frac{N_0}{N} \]
where \( R_0 \) is the count rate of the contemporary sample, \( R \) is the count rate of the ancient sample, \( N_0 \) is the initial number of nuclei, and \( N \) is the remaining number of nuclei.

### Step 2: Given values
- Count rate of contemporary wood, \( R_0 = 25.5 \, \text{cpm} \)
- Count rate of ancient wood, \( R = 20.5 \, \text{cpm} \)
- Half-life of \( ^{14}C \), \( t_{1/2} = 5770 \, \text{years} \)

### Step 3: Calculate the age of the ancient wood
Using the formula for age \( t \):
\[ t = \frac{t_{1/2}}{\ln(2)} \ln\left(\frac{R_0}{R}\right) \]
Substituting the known values:
\[ t = \frac{5770}{0.693} \ln\left(\frac{25.5}{20.5}\right) \]

### Step 4: Calculate the logarithm
First, calculate the ratio:
\[ \frac{25.5}{20.5} \approx 1.2439 \]
Now, calculate the natural logarithm:
\[ \ln(1.2439) \approx 0.219 \]

### Step 5: Substitute back to find age
Now substitute back into the age formula:
\[ t = \frac{5770}{0.693} \times 0.219 \]
Calculating this gives:
\[ t \approx 1817.5 \, \text{years} \]

### Step 6: Round to the nearest 50 years
Rounding 1817.5 years to the nearest 50 years gives:
\[ t \approx 1820 \, \text{years} \]

### Step 7: Calculate expected count rate for a 4000-year-old sample
We can use the same relationship to find the expected count rate \( R \) for a sample that is 4000 years old:
\[ t = \frac{t_{1/2}}{\ln(2)} \ln\left(\frac{R_0}{R}\right) \]
Rearranging gives:
\[ R = R_0 \cdot e^{-\frac{t \cdot \ln(2)}{t_{1/2}}} \]
Substituting \( t = 4000 \):
\[ R = 25.5 \cdot e^{-\frac{4000 \cdot 0.693}{5770}} \]

### Step 8: Calculate the exponential term
Calculating the exponent:
\[ -\frac{4000 \cdot 0.693}{5770} \approx -0.481 \]
Now calculate \( e^{-0.481} \approx 0.619 \).

### Step 9: Calculate the expected count rate
Now substituting back:
\[ R \approx 25.5 \cdot 0.619 \approx 15.77 \, \text{cpm} \]

### Final Answers
1. The age of the ancient wooden statue is approximately **1820 years**.
2. The expected count rate for a CO₂ sample that is 4000 years old is approximately **15.77 cpm**.