`"^(238)A_(92) to "^(234)B_(90)+"^4D_2+Q
In the given nuclear reaction ,the approximate amount of energy released will be :
[Given , mass of `"^(238)A_(92) = 238*05079 times 931.5MeV/c^2`,
mass of `_90^234B= 234*04363 times 931*5 MeV/c^2` ,
mass of `_2^4D = 4*00260 times931*5MeV/c^2]

Options

A. `3*82 MeV`
B. `5*9 MeV`
C. `4*25 MeV`
D. `2*12 MeV`

Answer

To solve the problem, we need to calculate the energy released in the nuclear reaction given by:

\[ ^{238}_{92}A \rightarrow ^{234}_{90}B + ^{4}_{2}D + Q \]

### Step 1: Calculate the mass defect (Δm)
We need to find the mass defect (Δm) in the reaction. The mass defect is the difference between the mass of the reactants and the mass of the products.

Given:
- Mass of \(^{238}_{92}A = 238.05079 \times 931.5 \, \text{MeV/c}^2\)
- Mass of \(^{234}_{90}B = 234.04363 \times 931.5 \, \text{MeV/c}^2\)
- Mass of \(^{4}_{2}D = 4.00260 \times 931.5 \, \text{MeV/c}^2\)

Now, we can calculate the mass defect:

\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} \]

\[ \Delta m = 238.05079 - (234.04363 + 4.00260) \]

Calculating the mass of the products:

\[ \text{Mass of products} = 234.04363 + 4.00260 = 238.04623 \]

Now substituting back into the equation for Δm:

\[ \Delta m = 238.05079 - 238.04623 = 0.00456 \, \text{u} \]

### Step 2: Convert mass defect to energy
The energy released (E) can be calculated using Einstein's mass-energy equivalence formula:

\[ E = \Delta m \times c^2 \]

In terms of MeV, we use the conversion factor:

\[ E = \Delta m \times 931.5 \, \text{MeV/u} \]

Substituting the value of Δm:

\[ E = 0.00456 \, \text{u} \times 931.5 \, \text{MeV/u} \]

Calculating the energy:

\[ E \approx 4.25 \, \text{MeV} \]

### Conclusion
The approximate amount of energy released in the reaction is:

\[ \boxed{4.25 \, \text{MeV}} \]