A radioactive isotope has a half life of T years. It radius to `3.125%` of its original value in

Options

A. 2T
B. 3T
C. 5T
D. 15T

Answer

To solve the problem of how long it takes for a radioactive isotope to reduce to `3.125%` of its original value, we can follow these steps:

### Step 1: Understand the percentage reduction
We know that `3.125%` can be expressed as a fraction:
\[ 3.125\% = \frac{3.125}{100} = 0.03125 \]

### Step 2: Set up the equation for radioactive decay
The number of radioactive nuclei remaining at time \( t \) can be expressed using the exponential decay formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N(t) \) is the number of nuclei at time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 3: Relate the decay constant to half-life
The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
In this case, since the half-life is given as \( T \), we have:
\[ \lambda = \frac{\ln(2)}{T} \]

### Step 4: Substitute the values into the decay equation
We want to find the time \( t \) when \( N(t) \) is \( 0.03125 N_0 \):
\[ 0.03125 N_0 = N_0 e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ 0.03125 = e^{-\lambda t} \]

### Step 5: Take the natural logarithm of both sides
Taking the natural logarithm gives us:
\[ \ln(0.03125) = -\lambda t \]

### Step 6: Substitute \( \lambda \) into the equation
Substituting \( \lambda = \frac{\ln(2)}{T} \):
\[ \ln(0.03125) = -\frac{\ln(2)}{T} t \]

### Step 7: Solve for \( t \)
Rearranging the equation to solve for \( t \):
\[ t = -\frac{T \ln(0.03125)}{\ln(2)} \]

### Step 8: Calculate \( \ln(0.03125) \)
We can calculate \( \ln(0.03125) \):
\[ \ln(0.03125) = \ln\left(\frac{1}{32}\right) = -\ln(32) = -\ln(2^5) = -5 \ln(2) \]
Substituting this back into the equation for \( t \):
\[ t = -\frac{T (-5 \ln(2))}{\ln(2)} = 5T \]

### Final Answer
Thus, the time \( t \) for the radioactive isotope to reduce to `3.125%` of its original value is:
\[ t = 5T \]

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