Given mass defect ( m_amu) = 0.05 a.m.u.

Step 1: Convert the mass defect from atomic mass units (a.m.u.) to kilograms (kg). Given mass defect ( m_amu) = 0.05 \, a.m.u. Given conversion factor: 1 \, a.m.u. = 1.67 × 10^-27 \, kg The mass defect in kilograms ( m) is: m = 0.05 \, a.m.u. × (1.67 × 10^-27 \, kg/a.m.u.) m = 0.0835 × 10^-27 \, kg m = 8.35 × 10^-29 \, kg Step 2: Use Einstein's mass-energy equivalence formula to calculate the energy released. The formula is E = m c^2, where: E is the energy released m is the mass defect in kilograms c is the speed of light in a vacuum Given speed of light (c) = 3.0 × 10^8 \, m/s Substitute the values into the formula: E = (8.35 × 10^-29 \, kg) × (3.0 × 10^8 \, m/s)^2 E = (8.35 × 10^-29) × (9.0 × 10^16) \, J E = (8.35 × 9.0) × (10^-29 × 10^16) \, J E = 75.15 × 10^(-29+16) \, J E = 75.15 × 10^-13 \, J Step 3: Express the energy in standard scientific notation. E = 7.515 × 10^-12 \, J The amount of energy released is 7.515 × 10^-12 J.