Number of `alpha`-particles emitted per second by a radioactive element falls to 1/32 of its original value in 50 days. The half-life-period of this element is

Options

A. 5 days
B. 15 days
C. 10 days
D. 20 days

Explanation

To determine the half-life period of a radioactive element given that the number of alpha particles emitted per second falls to \( \frac{1}{32} \) of its original value in 50 days, we can follow these steps:

### Step 1: Understand the relationship between the original value and the remaining value
We know that the number of alpha particles emitted falls to \( \frac{1}{32} \) of its original value. This can be expressed mathematically as:
\[ n = n_0 \left( \frac{1}{2} \right)^n \]
where:
- \( n \) is the remaining quantity,
- \( n_0 \) is the original quantity,
- \( n \) is the number of half-lives that have passed.

### Step 2: Set up the equation
From the information provided:
\[ \frac{n}{n_0} = \frac{1}{32} \]
This implies:
\[ \left( \frac{1}{2} \right)^n = \frac{1}{32} \]

### Step 3: Express \( \frac{1}{32} \) as a power of \( \frac{1}{2} \)
We can rewrite \( \frac{1}{32} \) as:
\[ \frac{1}{32} = \frac{1}{2^5} \]
Thus, we have:
\[ \left( \frac{1}{2} \right)^n = \left( \frac{1}{2} \right)^5 \]
From this, we can conclude:
\[ n = 5 \]

### Step 4: Relate the time to the number of half-lives
We know that the total time \( t \) is related to the half-life \( t_{1/2} \) and the number of half-lives \( n \) by the equation:
\[ t = n \cdot t_{1/2} \]
Given that \( t = 50 \) days and \( n = 5 \), we can substitute these values into the equation:
\[ 50 = 5 \cdot t_{1/2} \]

### Step 5: Solve for the half-life \( t_{1/2} \)
To find \( t_{1/2} \), we rearrange the equation:
\[ t_{1/2} = \frac{50}{5} = 10 \text{ days} \]

### Final Answer
The half-life period of the element is **10 days**.

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