A radioactive with half life `T=693.1` days. Emits `beta`-particles of average kinetic energy `E=8.4xx10^(-14)` joule. This radionuclide is used as source in a machine which generates electrical energy with efficiency `eta=12.6%`. Number of moles of the nuclide required to generate electrical energy at an initial rate is `P=441 KW`, is `nxx10^(n)` then find out value of `(n)/(m) (log_(e )2=0.6931)N_(A)=6.023xx10^(23)`

Explanation

To solve the problem step by step, we will follow the outlined reasoning from the video transcript and derive the necessary equations to find the required values.

### Step 1: Calculate the Decay Constant (λ)
The decay constant (λ) can be calculated using the half-life (T) of the radioactive substance:
\[ \lambda = \frac{\ln(2)}{T} \]
Given that \(T = 693.1\) days, we need to convert this into seconds:
\[ T = 693.1 \times 24 \times 3600 \text{ seconds} \]
Calculating this gives:
\[ T = 693.1 \times 86400 = 59719680 \text{ seconds} \]
Now substituting into the decay constant formula:
\[ \lambda = \frac{0.6931}{59719680} \approx 1.16 \times 10^{-9} \text{ s}^{-1} \]

### Step 2: Determine the Activity (A)
The activity (A) of the radioactive substance is given by:
\[ A = \lambda N \]
Where \(N\) is the number of atoms. The number of moles (n) is related to \(N\) by:
\[ N = n \times N_A \]
Where \(N_A = 6.023 \times 10^{23}\) is Avogadro's number. Thus, we can write:
\[ A = \lambda (n \times N_A) \]

### Step 3: Calculate the Power Input (P_input)
The power input to the machine can be expressed as:
\[ P_{\text{input}} = A \times E \]
Where \(E = 8.4 \times 10^{-14}\) joules is the average energy emitted per decay. Substituting for \(A\):
\[ P_{\text{input}} = \lambda (n \times N_A) \times E \]

### Step 4: Relate Output Power and Efficiency
The efficiency (η) is given as:
\[ \eta = \frac{P_{\text{output}}}{P_{\text{input}}} \]
Given \(P_{\text{output}} = 441 \text{ kW} = 441 \times 10^3 \text{ W}\), we can rearrange this to find \(P_{\text{input}}\):
\[ P_{\text{input}} = \frac{P_{\text{output}}}{\eta} = \frac{441 \times 10^3}{0.126} \]

Calculating this gives:
\[ P_{\text{input}} \approx 3500000 \text{ W} \text{ (or } 3.5 \times 10^6 \text{ W)} \]

### Step 5: Substitute and Solve for n
Now we can set the two expressions for \(P_{\text{input}}\) equal to each other:
\[ \lambda (n \times N_A) \times E = 3.5 \times 10^6 \]
Substituting the values we have:
\[ (1.16 \times 10^{-9}) (n \times 6.023 \times 10^{23}) (8.4 \times 10^{-14}) = 3.5 \times 10^6 \]
Calculating the left side:
\[ (1.16 \times 10^{-9}) (6.023 \times 10^{23}) (8.4 \times 10^{-14}) \approx 5.87 \times 10^{1} n \]
Setting this equal to \(3.5 \times 10^6\):
\[ 5.87 \times 10^{1} n = 3.5 \times 10^6 \]
Solving for \(n\):
\[ n \approx \frac{3.5 \times 10^6}{5.87 \times 10^{1}} \approx 5960 \approx 6 \times 10^3 \]

### Step 6: Find n and m
From the problem, we have \(n = 6\) and \(m = 3\) (since \(n \times 10^m\) implies \(m\) is the exponent).

### Step 7: Calculate the Ratio \(\frac{n}{m}\)
Finally, we need to find the ratio:
\[ \frac{n}{m} = \frac{6}{3} = 2 \]

### Final Answer
Thus, the value of \(\frac{n}{m}\) is:
\[ \boxed{2} \]