The energy released per fission of a 92U235 nucleus is nearly200eV20eV200MeV2000eV

Assuming that 200 MeV of energy is released per fission of 92U235 nucleus. Find the number of fission per second required to release 2 kW power3.125 × 10136.25 × 10133.125 × 10156.25 × 1014

Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 104 and 116. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be:

What is fired into a nucleus to cause nuclear fission?

Precise measurements give the following masses:nuclide mass23994Pu 239.0522u24194Pu 241.0569uparticle mass10n 1.0086649u0−1e ×5.485810−4u01e ×5.485810−4uNow consider the following nuclear reaction:→+23994Pu  210n  24194PuIs energy released or absorbed by this reaction?

About how much more energy is released by dividing a nucleus than by a molecule of TNT exploding?*1 point10^3 times7 times1,000 times10,000,000 times