The activity of a radioactive element decreases to one third of the original activity `I_(0)` in a period of nine years. After a further lapse of nine years, its activity will be

Options

A. `I_0`
B. `(2/3)I_0`
C. `(I_0/9)`
D. `(I_0/6)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial condition
We are given that the activity of a radioactive element decreases to one third of its original activity \( I_0 \) in a period of 9 years.

### Step 2: Write the equation for radioactive decay
The activity \( I \) at any time \( t \) can be described by the equation:
\[ I = I_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

### Step 3: Set up the equation for the first 9 years
At \( t = 9 \) years, the activity becomes:
\[ I = \frac{I_0}{3} \]
Substituting this into the decay equation gives:
\[ \frac{I_0}{3} = I_0 e^{-9\lambda} \]

### Step 4: Simplify the equation
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{3} = e^{-9\lambda} \]

### Step 5: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{3}\right) = -9\lambda \]
This can be rewritten as:
\[ -9\lambda = \ln(3^{-1}) = -\ln(3) \]
Thus, we find:
\[ \lambda = \frac{\ln(3)}{9} \]

### Step 6: Calculate the activity after a further 9 years
Now, we need to find the activity after a total of 18 years (9 years + another 9 years). We set \( t = 18 \):
\[ I' = I_0 e^{-18\lambda} \]
Substituting \( \lambda \):
\[ I' = I_0 e^{-18 \cdot \frac{\ln(3)}{9}} = I_0 e^{-2\ln(3)} \]

### Step 7: Simplify the expression
Using the property of exponents:
\[ e^{-2\ln(3)} = \left(e^{\ln(3)}\right)^{-2} = 3^{-2} = \frac{1}{9} \]
Thus, we have:
\[ I' = \frac{I_0}{9} \]

### Conclusion
After a further lapse of 9 years, the activity will be:
\[ I' = \frac{I_0}{9} \]

### Final Answer
The activity after 18 years will be \( \frac{I_0}{9} \).

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