Question

Calculate the quantity of energy produced per gram of U−235 (atomic mass = 235.043922 amu) for the neutron-induced fission of U−235 to form Xe−144 (atomic mass = 143.9385 amu) and Sr−90 (atomic mass =89.907738 amu).

Answer is absolute value E in J/g.

I think we're supposed to use E=mc^2?

Answer 1

The nuclear reaction is

U²³⁵ + n¹ Xe¹⁴⁴ + Sr ⁹⁰ + 2n¹

Δm = mass of reactant - mass of products

Δm = (235.043922 + 1.0086654 ) - ( 143.9385 + 89.907738 + 2 X 1.0086654) = 0.189019 amu

As 1 amu = 1.66054x10^-28 Kg

And Energy = mass X (speed of light)² = 0.189019 X 1.66054x10^-28 Kg X ( 3 X 10⁸ m/s)²

Energy = 2.825 X 10^-11 Joules

For one mole

Energy = 2.825 X 10^-11 Joules X Na = 2.825 X 10^-11 Joules X 6.023 X 10²³ J /mole

Therefore Energy / gram = Energy / mole / atomic weight of Uranium

Energy / g = 2.825 X 10^-11 Joules X 6.023 X 10²³ J /mole / 235 = 17.015 X 10¹² / 235 =7.24 X 10¹⁰ Joules /g