A `^7Li` target is bombarded with a proton beam current of `10^-4` A for 1 hour to produce `^Be` of activity `1.8xx10^8` disintegrations per second.
Assuming that `^7Be` radioactive nucleus is produced by bombarding 1000 protons, determine its half-life.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the total charge delivered by the proton beam
The current (I) is given as \(10^{-4}\) A, and the time (t) is 1 hour (3600 seconds). The total charge (Q) delivered can be calculated using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 10^{-4} \, \text{A} \times 3600 \, \text{s} = 3.6 \times 10^{-1} \, \text{C} \]

### Step 2: Calculate the number of protons bombarded
The charge of a single proton (e) is approximately \(1.6 \times 10^{-19}\) C. The number of protons (N) can be calculated using:
\[ N = \frac{Q}{e} \]
Substituting the values:
\[ N = \frac{3.6 \times 10^{-1} \, \text{C}}{1.6 \times 10^{-19} \, \text{C}} \approx 2.25 \times 10^{18} \, \text{protons} \]

### Step 3: Determine the number of \( ^7Be \) nuclei produced
According to the problem, it is assumed that 1000 protons produce 1 \( ^7Be \) nucleus. Therefore, the number of \( ^7Be \) nuclei (M) produced is:
\[ M = \frac{N}{1000} = \frac{2.25 \times 10^{18}}{1000} = 2.25 \times 10^{15} \, \text{nuclei} \]

### Step 4: Relate the activity of \( ^7Be \) to its decay constant
The activity (A) of the radioactive substance is given as \(1.8 \times 10^8\) disintegrations per second. The relationship between activity, decay constant (\( \lambda \)), and the number of nuclei is given by:
\[ A = \lambda M \]
Rearranging this gives us:
\[ \lambda = \frac{A}{M} \]
Substituting the values:
\[ \lambda = \frac{1.8 \times 10^8}{2.25 \times 10^{15}} \approx 8 \times 10^{-8} \, \text{s}^{-1} \]

### Step 5: Calculate the half-life of \( ^7Be \)
The half-life (\( t_{1/2} \)) is related to the decay constant by the formula:
\[ t_{1/2} = \frac{0.693}{\lambda} \]
Substituting the value of \( \lambda \):
\[ t_{1/2} = \frac{0.693}{8 \times 10^{-8}} \approx 8.66 \times 10^6 \, \text{s} \]

### Step 6: Convert the half-life from seconds to days
To convert seconds into days, we use the conversion:
\[ \text{days} = \frac{\text{seconds}}{86400} \quad (\text{since } 1 \text{ day} = 24 \times 60 \times 60 \text{ seconds}) \]
Calculating:
\[ \text{days} = \frac{8.66 \times 10^6}{86400} \approx 100.2 \, \text{days} \]

### Final Answer
The half-life of \( ^7Be \) is approximately **100.2 days**.

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