8g of the radioactive isotope, cesium -137 were collected on February 1 and kept in a sealed tube. On July 1, it was found that only 0.25 g of it remained. So the half-life period of the isotope is

Options

A. 37.5 days
B. 30 days
C. 23 days
D. 50 days

Explanation

To find the half-life period of the radioactive isotope cesium-137, we can follow these steps:

### Step 1: Determine the time period
Calculate the total time from February 1 to July 1.

- February has 28 days (since it is not a leap year).
- March has 31 days.
- April has 30 days.
- May has 31 days.
- June has 30 days.

Adding these up:
- February: 28 days
- March: 31 days
- April: 30 days
- May: 31 days
- June: 30 days

Total = 28 + 31 + 30 + 31 + 30 = 150 days

### Step 2: Determine the initial and remaining amounts
The initial amount of cesium-137 (A0) is 8 g, and the remaining amount (AD) after 150 days is 0.25 g.

### Step 3: Use the decay formula
The decay of a radioactive substance can be described by the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
Where:
- \( N(t) \) is the remaining quantity,
- \( N_0 \) is the initial quantity,
- \( \lambda \) is the decay constant,
- \( t \) is the time.

Alternatively, we can use the logarithmic form:
\[ \lambda = \frac{0.693}{T_{1/2}} \]
Where \( T_{1/2} \) is the half-life.

### Step 4: Calculate the decay constant
Using the logarithmic form:
\[ \lambda = \frac{2.303}{T} \log \left( \frac{A_0}{A_D} \right) \]
Substituting the known values:
- \( T = 150 \) days,
- \( A_0 = 8 \) g,
- \( A_D = 0.25 \) g.

Calculate:
\[ \lambda = \frac{2.303}{150} \log \left( \frac{8}{0.25} \right) \]
\[ \frac{8}{0.25} = 32 \]
\[ \log(32) \approx 1.5 \]

Thus:
\[ \lambda = \frac{2.303}{150} \times 1.5 \]

### Step 5: Calculate the half-life
Now, rearranging the decay constant formula to find the half-life:
\[ T_{1/2} = \frac{0.693}{\lambda} \]

Substituting the value of \( \lambda \):
\[ T_{1/2} = \frac{0.693}{\frac{2.303 \times 1.5}{150}} \]
Calculating this gives:
\[ T_{1/2} = \frac{0.693 \times 150}{2.303 \times 1.5} \]

Calculating the above expression:
1. \( 0.693 \times 150 \approx 103.95 \)
2. \( 2.303 \times 1.5 \approx 3.4545 \)
3. \( T_{1/2} \approx \frac{103.95}{3.4545} \approx 30 \) days.

### Final Answer
The half-life period of cesium-137 is approximately **30 days**.

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