Question

The mass defect for the following reaction, 1/1 H + 1/0n --> 2/1H , is 0.00239 amu. (1 amu= 1.66X10-27 kg) How much energy will this reaction provide?

) 2.39×10^−13J

b) 3.97×10^−13 J

c) 5.88×10^−13J

d) 7.21×10^−13J

Answer 1

The energy released in the given reaction is 3.97 × 10^-13 J.

Step-by-step explanation:

The mass defect for the given reaction is 0.00239 amu. To find the energy released, we need to convert the mass defect to kilograms and use the mass-energy equivalence equation, E=mc². Converting the mass defect to kg, we have (0.00239 amu) * (1.66 × 10^-27 kg/amu) = 3.97 × 10^-30 kg. Substituting this value into the equation, we get E = (3.97 × 10^-30 kg) * (299,792,458 m/s)² = 3.97 × 10^-13 J.