Question

For the reaction

X²⁰⁰→ A¹¹⁰ + B⁹⁰ + energy

if the BE/nucleon for the nucleus X²⁰⁰ is 7.4 MeV and for the nuclei A¹¹⁰ and B⁹⁰ is 8.2 MeV, then the energy released in the reaction is-

• A. 0.8 MeV
• B. 90 MeV
• C. 160 MeV
• D. 200 MeV

Answer 1

Answer: (C)

160 MeV

Answer 2

E= B.E_A + B.E_B – B.E_X