The present activity of the hair of Egyption mummy is 1.75 dpm `t_(1//2)` of `._(6)^(14)C` is 5770 year and disintegration rate of fresh smaple of `C^(14)` is 14 dpm . Find out age of mummy.

Options

A. 23080 year
B. 138480 year
C. 11998.3 year
D. 17310 year

Answer

To find the age of the Egyptian mummy using the activity of Carbon-14, we can follow these steps:

### Step 1: Understand the given data
- Present activity of the mummy's hair: \( A = 1.75 \, \text{dpm} \)
- Activity of fresh sample of Carbon-14: \( A_0 = 14 \, \text{dpm} \)
- Half-life of Carbon-14: \( t_{1/2} = 5770 \, \text{years} \)

### Step 2: Calculate the decay constant (\( \lambda \))
The decay constant (\( \lambda \)) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{\ln(2)}{5770} \]

### Step 3: Use the formula for radioactive decay
The relationship between the activities is given by:
\[ \frac{A}{A_0} = e^{-\lambda t} \]
Rearranging this gives:
\[ -\lambda t = \ln\left(\frac{A}{A_0}\right) \]
Thus,
\[ t = -\frac{\ln\left(\frac{A}{A_0}\right)}{\lambda} \]

### Step 4: Substitute the values into the equation
Substituting the activities:
\[ t = -\frac{\ln\left(\frac{1.75}{14}\right)}{\lambda} \]
Now, calculate \( \frac{1.75}{14} \):
\[ \frac{1.75}{14} = 0.125 \]
So,
\[ t = -\frac{\ln(0.125)}{\lambda} \]

### Step 5: Calculate \( \ln(0.125) \)
Using properties of logarithms:
\[ \ln(0.125) = \ln\left(\frac{1}{8}\right) = -\ln(8) = -3\ln(2) \]
Thus,
\[ t = -\frac{-3\ln(2)}{\lambda} = \frac{3\ln(2)}{\lambda} \]

### Step 6: Substitute \( \lambda \) back into the equation
Now substituting \( \lambda \):
\[ t = \frac{3\ln(2)}{\frac{\ln(2)}{5770}} = 3 \times 5770 \]

### Step 7: Calculate the age of the mummy
Calculating \( 3 \times 5770 \):
\[ t = 17310 \, \text{years} \]

### Final Answer
The age of the Egyptian mummy is approximately **17310 years**.

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