A sample of radioactive material decays simultaneouly 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 proecess B, and for further half an hour with both A and B. If, origianlly, there were `N_0` nuceli, 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`

Answer

To solve the problem step by step, we will analyze the decay processes A and B, calculate the remaining nuclei after each time interval, and finally combine the results.

### Step 1: Understanding the Decay Processes
- Process A has a half-life of \( T_{1/2} = \frac{1}{2} \) hours.
- Process B has a half-life of \( T_{1/2} = \frac{1}{4} \) hours.

### Step 2: First Half Hour (0 to 0.5 hours)
During the first half hour, only process A is active.
- Initial number of nuclei: \( N_0 \)
- After half an hour (1 half-life of A):
\[ N = N_0 \times \frac{1}{2} = \frac{N_0}{2} \]

### Step 3: Next One Hour (0.5 to 1.5 hours)
During this period, only process B is active.
- Remaining nuclei after the first half hour: \( \frac{N_0}{2} \)
- Number of half-lives of B in this hour:
\[ \text{Number of half-lives} = \frac{1 \text{ hour}}{\frac{1}{4} \text{ hour}} = 4 \]
- After one hour (4 half-lives of B):
\[ N = \frac{N_0}{2} \times \left(\frac{1}{2}\right)^4 = \frac{N_0}{2} \times \frac{1}{16} = \frac{N_0}{32} \]

### Step 4: Last Half Hour (1.5 to 2 hours)
During this period, both processes A and B are active.
- Remaining nuclei after the first 1.5 hours: \( \frac{N_0}{32} \)
- Effective half-life when both processes are active:
\[ T_{\text{effective}} = \frac{T_A \times T_B}{T_A + T_B} = \frac{\frac{1}{2} \times \frac{1}{4}}{\frac{1}{2} + \frac{1}{4}} = \frac{\frac{1}{8}}{\frac{3}{4}} = \frac{1}{6} \text{ hours} \]
- Number of half-lives in the last half hour:
\[ \text{Number of half-lives} = \frac{0.5 \text{ hours}}{\frac{1}{6} \text{ hours}} = 3 \]
- After the last half hour (3 half-lives):
\[ N = \frac{N_0}{32} \times \left(\frac{1}{2}\right)^3 = \frac{N_0}{32} \times \frac{1}{8} = \frac{N_0}{256} \]

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