A radioactive nucleus A with a half life T, decays into nucleus B. At t=0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3 . Then, t is given by

Options

A. `t = T log(1.3)`
B. `t = T/log(1.3)`
C. `t = (T/2)(log 2/ log1.3)`
D. `t = (T)(log 1.3/ log2)`

Answer

To solve the problem, we need to analyze the decay of a radioactive nucleus A into nucleus B and find the time \( t \) when the ratio of the number of B to that of A is 0.3.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
At \( t = 0 \), there is no nucleus B, which means:
\[ N_B(0) = 0 \quad \text{and} \quad N_A(0) = N_0 \]

2. **Expressing the Number of Nuclei**:
At any time \( t \), the total number of nuclei remains constant:
\[ N_A(t) + N_B(t) = N_0 \]
Given that at time \( t \), the ratio of the number of B to A is 0.3, we can express this as:
\[ N_B(t) = 0.3 N_A(t) \]

3. **Substituting into the Total Nuclei Equation**:
Substitute \( N_B(t) \) into the total nuclei equation:
\[ N_A(t) + 0.3 N_A(t) = N_0 \]
This simplifies to:
\[ 1.3 N_A(t) = N_0 \]
Therefore, we can express \( N_A(t) \) as:
\[ N_A(t) = \frac{N_0}{1.3} \]

4. **Applying the Radioactive Decay Law**:
The number of undecayed nuclei A at time \( t \) is given by the radioactive decay formula:
\[ N_A(t) = N_0 e^{-\lambda t} \]
Setting the two expressions for \( N_A(t) \) equal gives:
\[ \frac{N_0}{1.3} = N_0 e^{-\lambda t} \]

5. **Cancelling \( N_0 \)**:
Since \( N_0 \) is common on both sides, we can cancel it out (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{1.3} = e^{-\lambda t} \]

6. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{1.3}\right) = -\lambda t \]
This can be rewritten as:
\[ -\ln(1.3) = -\lambda t \]
Thus,
\[ t = \frac{\ln(1.3)}{\lambda} \]

7. **Relating Decay Constant to Half-Life**:
The decay constant \( \lambda \) is related to the half-life \( T \) by:
\[ \lambda = \frac{\ln(2)}{T} \]
Substituting this into the equation for \( t \):
\[ t = \frac{\ln(1.3)}{\frac{\ln(2)}{T}} = T \cdot \frac{\ln(1.3)}{\ln(2)} \]

8. **Final Result**:
Thus, the time \( t \) when the ratio of the number of B to that of A is 0.3 is given by:
\[ t = T \cdot \frac{\ln(1.3)}{\ln(2)} \]

### Conclusion:
The correct answer is:
\[ t = T \cdot \frac{\ln(1.3)}{\ln(2)} \]