The activity of a radioactive element reduces to `(1//16)th` of its original value in 30 years. Find its half life?

Options

A. 7.5 yrs
B. 75 yrs
C. 7.5 days
D. 75 month

Answer

To find the half-life of a radioactive element whose activity reduces to \( \frac{1}{16} \) of its original value in 30 years, we can follow these steps:

### Step 1: Understand the decay law
The decay law states that the number of active nuclei left after time \( t \) is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N(t) \) is the number of nuclei remaining after time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 2: Set up the equation
According to the problem, the activity reduces to \( \frac{1}{16} \) of its original value in 30 years. Thus, we can write:
\[ \frac{N(t)}{N_0} = \frac{1}{16} \]
Substituting this into the decay law gives:
\[ \frac{1}{16} N_0 = N_0 e^{-\lambda t} \]

### Step 3: Simplify the equation
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{16} = e^{-\lambda t} \]

### Step 4: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{16}\right) = -\lambda t \]

### Step 5: Express \( \frac{1}{16} \) in terms of powers of 2
We know that \( \frac{1}{16} = 2^{-4} \), so:
\[ \ln(2^{-4}) = -\lambda t \]
This simplifies to:
\[ -4 \ln(2) = -\lambda t \]

### Step 6: Solve for \( \lambda \)
Rearranging gives:
\[ \lambda = \frac{4 \ln(2)}{t} \]
Substituting \( t = 30 \) years:
\[ \lambda = \frac{4 \ln(2)}{30} \]

### Step 7: Use the half-life formula
The half-life \( T_{1/2} \) is given by:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting the expression for \( \lambda \):
\[ T_{1/2} = \frac{\ln(2)}{\frac{4 \ln(2)}{30}} = \frac{30}{4} = 7.5 \text{ years} \]

### Conclusion
Thus, the half-life of the radioactive element is \( 7.5 \) years.

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