The half-life of a radioactive substance is `30` minutes, The time (in minutes) taken between `40 %` decay and `85 %` decay of the same radioactive substance is.

Options

A. 45
B. 60
C. 15
D. 30

Explanation

To solve the problem of finding the time taken between 40% decay and 85% decay of a radioactive substance with a half-life of 30 minutes, we can follow these steps:

### Step 1: Understand the decay percentages
- The initial amount of the substance is denoted as \( N_0 \).
- After 40% decay, 60% of the substance remains, which means \( N = 0.6 N_0 \).
- After 85% decay, 15% of the substance remains, which means \( N = 0.15 N_0 \).

### Step 2: Use the decay formula
The decay of a radioactive substance can be described by the formula:
\[ N = N_0 e^{-\lambda t} \]
Where:
- \( N \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 3: Relate the decay to half-life
The decay constant \( \lambda \) can be expressed in terms of half-life \( T_{1/2} \):
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
Given that the half-life \( T_{1/2} = 30 \) minutes, we have:
\[ \lambda = \frac{\ln(2)}{30} \]

### Step 4: Calculate time for 40% decay
Let \( T_1 \) be the time taken to decay to 40%. We set up the equation:
\[ 0.6 N_0 = N_0 e^{-\lambda T_1} \]
Dividing both sides by \( N_0 \):
\[ 0.6 = e^{-\lambda T_1} \]
Taking the natural logarithm:
\[ \ln(0.6) = -\lambda T_1 \]
Substituting \( \lambda \):
\[ T_1 = -\frac{\ln(0.6)}{\lambda} = -\frac{\ln(0.6)}{\frac{\ln(2)}{30}} = -30 \frac{\ln(0.6)}{\ln(2)} \]

### Step 5: Calculate time for 85% decay
Let \( T_2 \) be the time taken to decay to 85%. We set up the equation:
\[ 0.15 N_0 = N_0 e^{-\lambda T_2} \]
Dividing both sides by \( N_0 \):
\[ 0.15 = e^{-\lambda T_2} \]
Taking the natural logarithm:
\[ \ln(0.15) = -\lambda T_2 \]
Substituting \( \lambda \):
\[ T_2 = -\frac{\ln(0.15)}{\lambda} = -\frac{\ln(0.15)}{\frac{\ln(2)}{30}} = -30 \frac{\ln(0.15)}{\ln(2)} \]

### Step 6: Find the time difference \( T_2 - T_1 \)
Now we can find the time taken between 40% decay and 85% decay:
\[ T_2 - T_1 = \left(-30 \frac{\ln(0.15)}{\ln(2)}\right) - \left(-30 \frac{\ln(0.6)}{\ln(2)}\right) \]
\[ = -30 \frac{\ln(0.15) - \ln(0.6)}{\ln(2)} \]
Using the property of logarithms:
\[ \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \]
\[ = -30 \frac{\ln\left(\frac{0.15}{0.6}\right)}{\ln(2)} = -30 \frac{\ln(0.25)}{\ln(2)} \]
Since \( \ln(0.25) = \ln\left(\frac{1}{4}\right) = -2\ln(2) \):
\[ = -30 \frac{-2\ln(2)}{\ln(2)} = 60 \text{ minutes} \]

### Final Answer
The time taken between 40% decay and 85% decay of the radioactive substance is **60 minutes**.

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