Question

In a nuclear reaction, 2 deuteron nuclei combine to form a helium nucleus. The energy released in MeV will be: (Given mass of deuteron = 2.01355 amu and mass of helium nucleus = 4.0028 amu.)

• A. 24.3 MeV
• B. 2.262 MeV
• C. 22.62 MeV
• D. 0.0243 MeV

Hint:

The energy released in a nuclear reaction can be calculated using the mass defect, which is the difference between the initial and final masses. This is then multiplied by 931.5 MeV (the energy equivalent of 1 amu) to obtain the energy release(D)

Answer 1

The Correct Option is C

The energy released in a nuclear reaction can be calculated using the mass defect. The mass defect is the difference in mass between the combined mass of the deuterons and the mass of the resulting helium nucleus. The energy corresponding to this mass defect is given by Einstein's equation: \[ E = \Delta m \cdot c^2 \] Where \( \Delta m \) is the mass defect, and \( c \) is the speed of light. First, calculate the mass defect: \[ \Delta m = (2 \times \text{mass of deuteron}) - \text{mass of helium nucleus} \] \[ \Delta m = 2 \times 2.01355 \, \text{amu} - 4.0028 \, \text{amu} = 4.0271 \, \text{amu} - 4.0028 \, \text{amu} = 0.0243 \, \text{amu} \] Next, convert the mass defect to energy. 1 amu is equivalent to 931.5 MeV, so the energy released is: \[ E = 0.0243 \, \text{amu} \times 931.5 \, \text{MeV/amu} = 22.62 \, \text{MeV} \] Thus, the energy released in the reaction is 22.62 MeV.