The energy released when 7 17.13 kg of ⁷₃ Li is converted into ⁴₂ He by proton bombardment is 10³² eV. The value of is _______. (Nearest integer) (Mass of ⁷₃ Li = 7.0183 u, mass of ⁴₂ He = 4.004 u, mass of proton = 1.008 u and 1 u = 931 MeV/c ^2 and Avogadro number = 6.0 10²³ )

Step-by-step solution

The nuclear reaction for the proton bombardment of Lithium is: ⁷₃ Li + ¹₁ H 2 ⁴₂ He The mass defect ( m ) for the reaction is: m = m(⁷₃ Li ) + m(¹₁ H ) - 2m(⁴₂ He ) m = 7.0183 + 1.008 - 2(4.004) m = 8.0263 - 8.008 = 0.0183 u Energy released per reaction ( Q ) is: Q = m 931 MeV = 0.0183 931 MeV = 17.0373 MeV Number of moles of ⁷₃ Li in 7 17.13 kg : n = 7 17.13 10^3 g 7 g/mol = 1000 17.13 mol Number of atoms of ⁷₃ Li : N = n N_A = 1000 17.13 6.0 10²³ = 6 17.13 10²⁶ Total energy released ( E ) is: E = N Q = ( 6 17.13