Question

4. In a nuclear plant, the final mass of the products is 6.32×10^-27kg, while the initial mass of the reactant is 6.30x10^-27kg, the energy released in the process is (speed of light in vacuum 3.0x10^8m/s, 1eV = 1.6x10^-19J) A.. 11.25meV B. 11.25 MJ C. 12.25MJ D. 12.25meV​

Answer 1

The energy released in the nuclear reaction is 3.70x10^10 millielectronvolts (meV).

Step-by-step explanation:

The energy released in a nuclear reaction can be calculated using Einstein's equation E = mc², where E is the energy released, m is the change in mass, and c is the speed of light in a vacuum. Given the final mass of the products (6.32×10^-27kg) and the initial mass of the reactant (6.30x10^-27kg), the change in mass is -0.02x10^-27kg.

Substituting the values into the equation, we can find the energy released:

E = (-0.02x10^-27kg)(3.0x10^8m/s)² = -5.93x10^12 J.

Converting the energy to electronvolts (eV), 1 J is equal to 6.24x10^18 eV. Therefore, the energy released is -5.93x10^12 J x (6.24x10^18 eV / 1 J) = -3.70x10^7 eV.

Converting from electronvolts (eV) to millielectronvolts (meV), 1 eV is equal to 10^3 meV. Therefore, the energy released is -3.70x10^7 eV x (10^3 meV / 1 eV) = -3.70x10^10 meV.

Since the energy released is negative, we take the absolute value to get the final answer: 3.70x10^10 meV.