Question

A slow neutron strikes a nucleus of ${}_{92}^{235}U$ splitting it into lighter nuclei of ${}_{56}^{141}Ba$ and ${}_{36}^{92}Kr$ along with three neutrons. The energy released in this reaction is :
(The masses of Uranium, Barium and Krypton in this reaction are $235.043933$ a.m.u, $140.917700$ a.m.u and $91.895400$ a.m.u respectively. The mass of a neutron is $1.008665$ a.m.u)

• A. $198.9$ MeV
• B. $156.9$ MeV
• C. $186.9$ MeV
• D. $209.8$ MeV

Answer 1

The correct option is A $198.9$ MeV

${}_{92}^{235}U+{}_0^{1}n\to {}_{56}^{141}Ba+{}_{36}^{92}Kr+3\left({}_0^{1}n\right)+Q$

Q is Energy released Q = ( Total mass of reactants – total mass ) $\times C^2$

$=\left[Mass\left({}_{92}^{235}U\right)+mass\left({}_0^{1}n\right)\right]-\left[mass{}_{56}^{141}Ba+mass{}_{36}^{92}Kr+mass3{}_0^{1}n\right]c^2$

$=\left(235\cdot 043933-140.917700-91.8954-2\times 1.008665\right)u\times c^2$

$=198.9MeV$