Question

How much mass defect is required to release 5.5 x 10²⁰ J of energy?

A reaction takes place with a mass defect of 0.030kg. How much energy does this correspond to?

A reaction of 2.5kg of fissionable material takes place. Assuming a 0.10% mass defect, how much energy is released?

Answer 1

1) To Convert joule into eV one has to consider the factor 6.2 * 10¹⁸

1 J = 6.2 * 10¹⁸ eV

Thus 5.5 *10²⁰ J = 6.2* 5.5 * 10¹⁸⁺²⁰= 34.1 * 10³⁸ eV = 34.1 * 10³² MeV

According to mass defect relation:

E= MeV

34.1 * 10³² MeV = 931.5 *

= 3.6 * 10³⁰ amu

2) 0.030 kg = 1.80 * 10²⁵ amu.

E= MeV

= 931.5 * 1.8* 10²⁵ MeV

= 1676.7 * 10²⁵ MeV.

3) 0.10% of 2.5kg = 0.25kg ( mass defect)

0.25 kg = 1.5 * 10²⁶ amu

E= MeV

= 931.5 * 1.5 * 10²⁶ MeV

⁼1397.25 * 10²⁶ MeV