The `._(92)U^(235)` absorbs a slow neturon (thermal neutron) & undergoes a fission represented by `._(92)U^(235)+._(0)n^(1)rarr._(92)U^(236)rarr._(56)Ba^(141)+_(36)Kr^(92)+3_(0)n^(1)+E`. Calculate:
The energy released when `1 g` of `._(92)U^(235)` undergoes complete fission in `N` if `m=[N]` then find `(m-2)/(5)`. `[N]` greatest integer
Given
`._(92)U^(235)=235.1175"amu (atom)"`,
`._(56)Ba^(141)=140.9577 "amu (atom)"`,
`._(36)r^(92)=91.9263 "amu(atom)" , ._(0)n^(1)=1.00898 "amu", 1 "amu"=931 MeV//C^(2)`

Explanation

To solve the problem, we need to calculate the energy released when 1 gram of Uranium-235 undergoes complete fission. We will follow these steps:

### Step 1: Write down the reaction and the masses involved
The fission reaction is:
\[ _{92}^{235}U + _{0}^{1}n \rightarrow _{56}^{141}Ba + _{36}^{92}Kr + 3 \times _{0}^{1}n + E \]
The given masses are:
- Mass of \( _{92}^{235}U = 235.1175 \, \text{amu} \)
- Mass of \( _{56}^{141}Ba = 140.9577 \, \text{amu} \)
- Mass of \( _{36}^{92}Kr = 91.9263 \, \text{amu} \)
- Mass of neutron \( _{0}^{1}n = 1.00898 \, \text{amu} \)

### Step 2: Calculate the mass defect
The mass defect (\(\Delta m\)) can be calculated using the formula:
\[ \Delta m = \text{mass of reactants} - \text{mass of products} \]
The mass of reactants is:
\[ \text{mass of } U + \text{mass of neutron} = 235.1175 + 1.00898 = 236.12648 \, \text{amu} \]
The mass of products is:
\[ \text{mass of } Ba + \text{mass of } Kr + 3 \times \text{mass of neutron} = 140.9577 + 91.9263 + 3 \times 1.00898 \]
Calculating the mass of products:
\[ = 140.9577 + 91.9263 + 3.02694 = 235.91094 \, \text{amu} \]
Now, calculate the mass defect:
\[ \Delta m = 236.12648 - 235.91094 = 0.21554 \, \text{amu} \]

### Step 3: Calculate the energy released
The energy released (\(E\)) can be calculated using the mass-energy equivalence:
\[ E = \Delta m \times 931 \, \text{MeV/amu} \]
Substituting the values:
\[ E = 0.21554 \times 931 \approx 200.57 \, \text{MeV} \]

### Step 4: Calculate the number of fissions in 1 gram of \( _{92}^{235}U \)
The number of moles of \( _{92}^{235}U \) in 1 gram is given by:
\[ n = \frac{1 \, \text{g}}{235.1175 \, \text{g/mol}} \approx 0.00425 \, \text{mol} \]
Using Avogadro's number (\(N_A \approx 6.022 \times 10^{23} \, \text{atoms/mol}\)), the number of atoms in 1 gram of \( _{92}^{235}U \) is:
\[ N = n \times N_A \approx 0.00425 \times 6.022 \times 10^{23} \approx 2.56 \times 10^{21} \, \text{atoms} \]

### Step 5: Calculate the total energy released
The total energy released when 1 gram of \( _{92}^{235}U \) undergoes complete fission is:
\[ \text{Total Energy} = N \times E \approx 2.56 \times 10^{21} \times 200.57 \, \text{MeV} \]
Convert MeV to Joules (1 MeV = \(1.602 \times 10^{-13} \, \text{J}\)):
\[ \text{Total Energy} \approx 2.56 \times 10^{21} \times 200.57 \times 1.602 \times 10^{-13} \approx 8.20 \times 10^{6} \, \text{J} \]

### Step 6: Convert to Megawatt-hours
To convert Joules to Megawatt-hours:
\[ \text{Energy in MWh} = \frac{8.20 \times 10^{6}}{3.6 \times 10^{6}} \approx 2.28 \, \text{MWh} \]

### Step 7: Find \(N\) and \(m\)
Since \(N\) represents the energy released in MWh, we take the greatest integer:
\[ N \approx 2 \]
Thus, \(m = N\).

### Step 8: Calculate \((m-2)/5\)
\[ \frac{m-2}{5} = \frac{2-2}{5} = 0 \]

### Final Answer
The final answer is:
\[ \frac{m-2}{5} = 0 \]