If `N_(0) and N` are the number of radioactive particles at time `t = 0 and t = t`, then

Options

A. `lamda = (1)/(t) "log" (N_(0))/(N)`
B. `lamda = (2.303)/(t) "log"(N)/(N_(0))`
C. `lamda = (t)/(2.303)"log"(N_(0))/(N)`
D. `lamda = (2.303)/(t) "log" (N_(0))/(N)`

Answer

To solve the problem regarding the relationship between the number of radioactive particles at two different times, we can use the concept of radioactive decay. The number of radioactive particles decreases over time according to the decay constant (λ).

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( N_0 \) be the number of radioactive particles at time \( t = 0 \).
- Let \( N \) be the number of radioactive particles at time \( t \).
- \( \lambda \) is the decay constant.

2. **Radioactive Decay Formula**:
The number of radioactive particles remaining after time \( t \) can be expressed using the radioactive decay formula:
\[ N = N_0 e^{-\lambda t} \]

3. **Taking Logarithms**:
To express this in a logarithmic form, we can rearrange the equation:
\[ \frac{N}{N_0} = e^{-\lambda t} \]
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{N}{N_0}\right) = -\lambda t \]

4. **Rearranging for Decay Constant**:
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = -\frac{1}{t} \ln\left(\frac{N}{N_0}\right) \]

5. **Using Logarithm Base 10**:
If we want to express this in terms of base 10 logarithms, we can use the conversion:
\[ \ln(x) = 2.303 \log_{10}(x) \]
Therefore, we can rewrite the decay constant as:
\[ \lambda = -\frac{2.303}{t} \log_{10}\left(\frac{N}{N_0}\right) \]

6. **Final Expression**:
Rearranging gives us:
\[ \lambda = \frac{2.303}{t} \log_{10}\left(\frac{N_0}{N}\right) \]

### Conclusion:
The final expression for the decay constant \( \lambda \) in terms of the initial and remaining number of radioactive particles is:
\[ \lambda = \frac{2.303}{t} \log_{10}\left(\frac{N_0}{N}\right) \]