Consider the fission by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ¹⁴⁰₅₈Ce and ⁹⁹₄₄Ru. Calculate Q for this fission process. The relevant atomic and particle masses are

m(²³⁸₉₂U) = 238.05079 u

m(¹⁴⁰₅₈Ce) = 139.90543 u

m(⁹⁹₄₄Ru) = 98.90594 u

Given that,

Mass of a nucleus , m₁ = 238.05079 u

Mass of a nucleus , m₂ = 139.90543 u

Mass of a nucleus , m₃ = 98.90594 u

Mass of a neutron , m₄ = 1.008665 u

In the fission of , 10 beta particles are emitted. The nuclear reaction is,

Energy released during above nuclear fission is given by,

Where,

m’ represents the corresponding atomic masses of the nuclei

= m₁ – 92m_e (∵ atomic number 92)

= m₂ – 58m_e (∵ atomic number 58)

= m₃ – 44m_e (∵ atomic number 44)

= m₄

∴

=

= (0.247995)u×c²

But, u = 931.5 MeV/c²

∴ Q = 0.247995×931.5 Mev

= 231.007 MeV

The energy released during given fission reaction is 231.007 MeV.