The half-life of a radioactive element which has only `(1)/(32)` of its original mass left after a lapse of `60` days is

Options

A. 12 days
B. 32 days
C. 60 days
D. 64 days

Explanation

To find the half-life of a radioactive element that has only \( \frac{1}{32} \) of its original mass left after 60 days, we can use the concept of half-life in radioactive decay. Here’s a step-by-step solution:

### Step 1: Understand the decay process
The amount of a radioactive substance remaining after a certain time can be expressed using the formula:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{T}{t_{1/2}}} \]
where:
- \( N \) is the remaining quantity of the substance,
- \( N_0 \) is the initial quantity,
- \( T \) is the elapsed time,
- \( t_{1/2} \) is the half-life of the substance.

### Step 2: Set up the equation
Given that \( N = \frac{N_0}{32} \) and \( T = 60 \) days, we can substitute these values into the equation:
\[ \frac{N_0}{32} = N_0 \left( \frac{1}{2} \right)^{\frac{60}{t_{1/2}}} \]

### Step 3: Simplify the equation
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{32} = \left( \frac{1}{2} \right)^{\frac{60}{t_{1/2}}} \]

### Step 4: Express \( \frac{1}{32} \) as a power of \( \frac{1}{2} \)
Recognizing that \( \frac{1}{32} = \frac{1}{2^5} \), we can rewrite the equation:
\[ \frac{1}{2^5} = \left( \frac{1}{2} \right)^{\frac{60}{t_{1/2}}} \]

### Step 5: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ 5 = \frac{60}{t_{1/2}} \]

### Step 6: Solve for \( t_{1/2} \)
To find \( t_{1/2} \), we can rearrange the equation:
\[ t_{1/2} = \frac{60}{5} = 12 \text{ days} \]

### Conclusion
Thus, the half-life of the radioactive element is \( 12 \) days.

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