A sample of a radioactive substance undergoes 80% decomposition in 345 minutes. Its half life is …………minutes

Options

A. `(In2)/(In5)xx345`
B. `(In5)/(In2)xx345`
C. `(In5)/(In4)xx345`
D. `(In4)/(In5)xx345`

Answer

To find the half-life of a radioactive substance that undergoes 80% decomposition in 345 minutes, we can follow these steps:

### Step 1: Understand the decomposition
If a sample undergoes 80% decomposition, it means that 20% of the original sample remains.

### Step 2: Set up the equation
Let the initial amount of the substance be \( N_0 \). After 345 minutes, the remaining amount \( N \) can be expressed as:
\[ N = N_0 \times (1 - 0.80) = N_0 \times 0.20 \]

### Step 3: Use the radioactive decay formula
The radioactive decay can be described by the equation:
\[ N = N_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant and \( t \) is the time in minutes.

### Step 4: Substitute the known values
Substituting the values into the equation gives:
\[ 0.20 N_0 = N_0 e^{-\lambda \cdot 345} \]
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ 0.20 = e^{-\lambda \cdot 345} \]

### Step 5: Take the natural logarithm of both sides
Taking the natural logarithm on both sides:
\[ \ln(0.20) = -\lambda \cdot 345 \]

### Step 6: Solve for the decay constant \( \lambda \)
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = -\frac{\ln(0.20)}{345} \]

### Step 7: Relate the decay constant to half-life
The half-life \( t_{1/2} \) is related to the decay constant \( \lambda \) by the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]

### Step 8: Substitute \( \lambda \) into the half-life equation
Substituting the expression for \( \lambda \):
\[ t_{1/2} = \frac{\ln(2)}{-\frac{\ln(0.20)}{345}} = \frac{345 \ln(2)}{-\ln(0.20)} \]

### Step 9: Simplify the expression
Using the property of logarithms, we can express \( \ln(0.20) \) as:
\[ \ln(0.20) = \ln\left(\frac{1}{5}\right) = -\ln(5) \]
Thus, we have:
\[ t_{1/2} = \frac{345 \ln(2)}{\ln(5)} \]

### Final Calculation
Now, we can calculate the numerical value of \( t_{1/2} \) using the natural logarithm values:
- \( \ln(2) \approx 0.693 \)
- \( \ln(5) \approx 1.609 \)

Substituting these values:
\[ t_{1/2} \approx \frac{345 \times 0.693}{1.609} \approx 150.4 \text{ minutes} \]

### Conclusion
The half-life of the radioactive substance is approximately **150.4 minutes**.

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