Radioactivity of a radioactive element remains `1//10` of the original radioactivity after `2.303` seconds. The half life period is

Options

A. 2.303sec
B. 0.2303sec
C. 0.693sec
D. 0.0693sec

Answer

To find the half-life period of a radioactive element given that its radioactivity remains \( \frac{1}{10} \) of the original radioactivity after \( 2.303 \) seconds, we can follow these steps:

### Step 1: Understand the relationship between original and remaining radioactivity
Let \( A_0 \) be the original radioactivity. After \( 2.303 \) seconds, the remaining radioactivity \( A_t \) is given as:
\[ A_t = \frac{A_0}{10} \]

### Step 2: Use the first-order kinetics equation
For a first-order reaction, the relationship between the remaining amount and the original amount can be expressed as:
\[ A_t = A_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Where \( t \) is the time elapsed and \( t_{1/2} \) is the half-life period.

### Step 3: Substitute the known values into the equation
Substituting \( A_t \) and \( t \) into the equation:
\[ \frac{A_0}{10} = A_0 \left( \frac{1}{2} \right)^{\frac{2.303}{t_{1/2}}} \]
We can cancel \( A_0 \) from both sides (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{10} = \left( \frac{1}{2} \right)^{\frac{2.303}{t_{1/2}}} \]

### Step 4: Take logarithm of both sides
Taking logarithm base 10 on both sides:
\[ \log_{10} \left( \frac{1}{10} \right) = \frac{2.303}{t_{1/2}} \log_{10} \left( \frac{1}{2} \right) \]
Since \( \log_{10} \left( \frac{1}{10} \right) = -1 \) and \( \log_{10} \left( \frac{1}{2} \right) = -\log_{10}(2) \):
\[ -1 = \frac{2.303}{t_{1/2}} (-\log_{10}(2)) \]

### Step 5: Rearrange the equation to solve for \( t_{1/2} \)
Rearranging gives:
\[ 1 = \frac{2.303 \log_{10}(2)}{t_{1/2}} \]
Thus,
\[ t_{1/2} = 2.303 \log_{10}(2) \]

### Step 6: Substitute the value of \( \log_{10}(2) \)
Using \( \log_{10}(2) \approx 0.301 \):
\[ t_{1/2} = 2.303 \times 0.301 \]

### Step 7: Calculate \( t_{1/2} \)
Calculating the above gives:
\[ t_{1/2} \approx 0.693 \text{ seconds} \]

### Final Answer
The half-life period \( t_{1/2} \) is \( 0.693 \) seconds.

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