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 that reduces to \( \frac{1}{16} \) of its original activity in 30 years, we can follow these steps:

### Step 1: Understand the relationship between activity and half-life
The activity of a radioactive substance decreases over time. The half-life (\( T_{1/2} \)) is the time taken for the activity to reduce to half of its original value.

### Step 2: Set up the equation based on the problem statement
Let the initial activity be \( A_0 \). After 30 years, the activity is given as:
\[ A = \frac{A_0}{16} \]
This means that in 30 years, the activity has reduced to \( \frac{1}{16} \) of its original value.

### Step 3: Determine how many half-lives have passed
To find how many half-lives have occurred in 30 years, we can express \( \frac{A_0}{16} \) in terms of half-lives:
\[ \frac{A_0}{16} = \frac{A_0}{2^n} \]
where \( n \) is the number of half-lives.

Since \( \frac{1}{16} = \frac{1}{2^4} \), we can conclude:
\[ n = 4 \]
This means that 4 half-lives have passed in 30 years.

### Step 4: Calculate the half-life
Now, we can find the duration of one half-life:
\[ \text{Total time} = n \times T_{1/2} \]
Substituting the known values:
\[ 30 \text{ years} = 4 \times T_{1/2} \]
To find \( T_{1/2} \):
\[ T_{1/2} = \frac{30 \text{ years}}{4} = 7.5 \text{ years} \]

### Final Answer
The half-life of the radioactive element is \( 7.5 \) years.

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