QJEEM23Ph5c98: JEE Main 2023 Physics - Class 12 JEEM23Ph5c98 Solution: In the given nuclear reaction, the approximate amount of energy …

The energy released in a nuclear reaction can be determined by the mass difference between the reactants and the products, multiplied by the speed of light squared, as per Einstein's mass-energy equivalence relation, $E=mc^2$ .

In the given nuclear reaction, the energy released is:

$Q = \left( m_{A} - m_{B} - m_{D} \right) \times 931.5 \ \text{MeV/c}^2$

We are given the masses of A, B, and D as follows:

$m_{A} = 238.05079 \times 931.5 \ \text{MeV/c}^2$

$m_{B} = 234.04363 \times 931.5 \ \text{MeV/c}^2$

$m_{D} = 4.00260 \times 931.5 \ \text{MeV/c}^2$

Now, we can substitute these values into the equation for Q:

$Q = \left( (238.05079 - 234.04363 - 4.00260) \times 931.5 \right) \ \text{MeV}$

$Q = \left( (0.00456) \times 931.5 \right) \ \text{MeV}$

Calculating the energy released:

$Q \approx 4.25 \ \text{MeV}$

So, the approximate amount of energy released in the given nuclear reaction is 4.25 MeV.