A certain radioactive element has a half-life of 20 years . If we have a block with 10 g of the element in it, after how many years will there be just 2.5 gm of element in the block

Options

A. 80 years
B. 40 years
C. 100 years
D. 60 years

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given information
- Initial amount of the radioactive element, \( n_0 = 10 \) g
- Final amount of the radioactive element, \( n = 2.5 \) g
- Half-life of the element, \( T_{1/2} = 20 \) years

### Step 2: Use the half-life formula to find the decay constant (\( \lambda \))
The relationship between half-life and decay constant is given by the formula:
\[ T_{1/2} = \frac{0.693}{\lambda} \]
Rearranging this to find \( \lambda \):
\[ \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{20} \approx 0.03465 \, \text{years}^{-1} \]

### Step 3: Use the exponential decay formula
The amount of radioactive substance remaining after time \( t \) can be calculated using the formula:
\[ n = n_0 e^{-\lambda t} \]
Substituting the known values:
\[ 2.5 = 10 e^{-0.03465 t} \]

### Step 4: Solve for \( e^{-0.03465 t} \)
Dividing both sides by 10:
\[ 0.25 = e^{-0.03465 t} \]

### Step 5: Take the natural logarithm of both sides
\[ \ln(0.25) = -0.03465 t \]

### Step 6: Solve for \( t \)
Calculating \( \ln(0.25) \):
\[ \ln(0.25) = \ln\left(\frac{1}{4}\right) = -\ln(4) \approx -1.3863 \]
Now substituting this value back:
\[ -1.3863 = -0.03465 t \]
Thus,
\[ t = \frac{1.3863}{0.03465} \approx 40 \text{ years} \]

### Conclusion
After 40 years, there will be just 2.5 g of the radioactive element remaining in the block.

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