A sample of radioactive material decays simultaneously by two processes A and B with half-lives `(1)/(2)` and `(1)/(4)h`, respectively. For the first half hour it decays with the process A, next one hour with the process B, and for further half an hour with both A and B. If, originally, there were `N_0` nuclei, find the number of nuclei after 2 h of such decay.

Options

A. `N_0/(2)^(8)`
B. `N_0/(2)^(4)`
C. `N_0/(2)^(6)`
D. `N_0/(2)^(5)`

Explanation

To solve the problem of radioactive decay of a sample with two processes A and B, we will follow these steps:

### Step 1: Understand the decay processes and their half-lives
- Process A has a half-life of \( \frac{1}{2} \) hour.
- Process B has a half-life of \( \frac{1}{4} \) hour.

### Step 2: Calculate the decay for the first half hour (0 to 0.5 hours)
During the first half hour, only process A is active. Since the half-life of A is \( \frac{1}{2} \) hour, after this time:
\[ N = N_0 \times \frac{1}{2} \]
Thus, the number of nuclei after the first half hour is:
\[ N = \frac{N_0}{2} \]

### Step 3: Calculate the decay for the next one hour (0.5 to 1.5 hours)
In this interval, only process B is active. The half-life of B is \( \frac{1}{4} \) hour. In one hour, the number of half-lives for B is:
\[ \text{Number of half-lives} = \frac{1 \text{ hour}}{\frac{1}{4} \text{ hour}} = 4 \]
The number of nuclei after this hour is given by:
\[ N = N \times \left(\frac{1}{2}\right)^4 \]
Substituting the value of N from the previous step:
\[ N = \frac{N_0}{2} \times \left(\frac{1}{2}\right)^4 = \frac{N_0}{2} \times \frac{1}{16} = \frac{N_0}{32} \]

### Step 4: Calculate the decay for the last half hour (1.5 to 2 hours)
In this interval, both processes A and B are active. We need to find the effective half-life when both processes are active. The effective half-life \( T_{eff} \) can be calculated using the formula:
\[ \frac{1}{T_{eff}} = \frac{1}{T_A} + \frac{1}{T_B} \]
Where:
- \( T_A = \frac{1}{2} \) hour
- \( T_B = \frac{1}{4} \) hour

Calculating \( T_{eff} \):
\[ \frac{1}{T_{eff}} = \frac{1}{\frac{1}{2}} + \frac{1}{\frac{1}{4}} = 2 + 4 = 6 \]
Thus, \( T_{eff} = \frac{1}{6} \) hour.

Now, in the half hour (0.5 hours), the number of half-lives for the effective decay is:
\[ \text{Number of half-lives} = \frac{\frac{1}{2}}{\frac{1}{6}} = 3 \]
Now, applying this to the number of nuclei:
\[ N = N \times \left(\frac{1}{2}\right)^3 \]
Substituting the value of N from the previous step:
\[ N = \frac{N_0}{32} \times \left(\frac{1}{2}\right)^3 = \frac{N_0}{32} \times \frac{1}{8} = \frac{N_0}{256} \]

### Final Result
After 2 hours, the number of nuclei remaining is:
\[ N = \frac{N_0}{256} \]