4. The amount of energy released in fission of 235gm of U-235 is * 1 O (a) 1.926 x 10^13 Mev O (6) 1.926 x 10^13 J O (C) 1.926 x 10^23 Mev O (d) 1.926 x 10^23 J
Answers
we have to find the amount of energy released in fission of 235 gm of U-235 is ...
solution : the fission process represented by the equation,
₉₂U²³⁵ + ₀n¹ ⇒ ₅₆Ba¹⁴⁴ + ₃₆Kr⁸⁹ + 3₀n¹
so the mass of reactant = mass of uranium + mass of neutron
= 234.39 amu + 1.01 amu = 235.40 amu
the mass of product = mass of Ba + mass of Kr + 3 * mass of neutron
= 143.28 amu + 88.89 amu + 3(1.01 amu) = 235.2 amu
now change in mass = mass of reactant - mass of product
= 235.4 - 235.2 = 0.2 amu
as 1 amu = 931 MeV
so, 0.2 amu = 186.2 MeV
here we get, a single atom of U-235 released 200MeV (approx) energy.
now, no of atoms in 235gm of U-235 = no of moles × avogadro's number
= 235/235 × 6.023 × 10²³
= 6.023 × 10²³
now the amount of energy released in fission of 235g of U-235 is 6.023 × 10²³ × 186.2 MeV = 1121.4826 × 10²³ MeV
1 MeV = 1.6 × 10⁻¹³ J
so, 11214.826 × 10²³ MeV = 1121.4826 × 10²³ × 1.6 × 10⁻¹³ = 17943.7216 × 10¹⁰
= 1.7943 × 10¹³ J this is the nearest value of 1.926 × 10¹³ J
so the amount of energy released in fission of 235 gm of U -235 is 1.926 × 10¹³ J