A radioactive sample decays through two different decay processes `alpha`-decay and `beta`-decay. Half-life time for `alpha`−decay is 3h and half-life time for `beta`−decay is `6h`. What will be the ratio of number of initial radioactive to the number of nuclei present after `6 h`?.

Explanation

To solve the problem, we need to find the ratio of the initial number of radioactive nuclei to the number of nuclei present after 6 hours, considering both alpha and beta decay processes.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Half-life for alpha decay, \( T_{\alpha} = 3 \) hours
- Half-life for beta decay, \( T_{\beta} = 6 \) hours
- Time, \( t = 6 \) hours

2. **Calculate the Decay Constants:**
The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

- For alpha decay:
\[ \lambda_{\alpha} = \frac{\ln(2)}{3 \text{ hours}} \]
- For beta decay:
\[ \lambda_{\beta} = \frac{\ln(2)}{6 \text{ hours}} \]

3. **Calculate the Effective Decay Constant:**
The effective decay constant \( \lambda_{\text{effective}} \) for both processes is given by:
\[ \lambda_{\text{effective}} = \lambda_{\alpha} + \lambda_{\beta} \]

Substituting the values:
\[ \lambda_{\text{effective}} = \frac{\ln(2)}{3} + \frac{\ln(2)}{6} \]

To add these fractions, we need a common denominator:
\[ \lambda_{\text{effective}} = \frac{2\ln(2)}{6} + \frac{\ln(2)}{6} = \frac{3\ln(2)}{6} = \frac{\ln(2)}{2} \]

4. **Use the Decay Law:**
The number of nuclei remaining after time \( t \) is given by:
\[ N = N_0 e^{-\lambda_{\text{effective}} t} \]

Substituting \( t = 6 \) hours:
\[ N = N_0 e^{-\left(\frac{\ln(2)}{2}\right)(6)} = N_0 e^{-3\ln(2)} = N_0 \left(e^{\ln(2)}\right)^{-3} = N_0 \left(\frac{1}{2}\right)^3 = N_0 \cdot \frac{1}{8} \]

5. **Calculate the Ratio:**
The ratio of the initial number of radioactive nuclei \( N_0 \) to the number of nuclei present after 6 hours \( N \) is:
\[ \text{Ratio} = \frac{N_0}{N} = \frac{N_0}{\frac{N_0}{8}} = 8 \]

### Final Answer:
The ratio of the number of initial radioactive nuclei to the number of nuclei present after 6 hours is **8**.

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