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 235.0439 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

Hint:The nuclear fission is a process in which an atom splits up into two or more particles and a huge amount of energy is released and there are some by-products particles. The nuclear fission is an example of conversion of mass into energy.

Complete step by step answer:
It is given in the problem that 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 and we need to find the energy released during this process.
This is a nuclear fission reaction as a particle is split into two particles and energy is released.
The reaction that takes place in this nuclear fission is given by,
$ \Rightarrow {}_{92}^{235}U{ + ^1}n \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + {3^1}n$
Change in mass is equal to,
$ \Rightarrow \Delta m = {m_U} - {m_{Ba}} - {m_{kr}} - 2{m_n}$
As it is given that the mass of the Uranium is 235.0439 a.m.u, the mass of the barium is 140.917700 a.m.u and the mass of the neutron is 1.008665 a.m.u.
Replace the mass of the uranium the mass of Barium the mass of krypton and mass of neutron.
$ \Rightarrow \Delta m = 235 \cdot 043933 - 140 \cdot 9177 - 91 \cdot 8954 - 2\left( {1 \cdot 008665} \right)$
$ \Rightarrow \Delta m = 0 \cdot 213503\;{\text{a}}{\text{.m}}{\text{.u}}$
$ \Rightarrow E = \Delta m{c^2}$
Where ${c^2}$ in MeV is equal to 931.49432MeV.
$ \Rightarrow E = \Delta m{c^2}$
$ \Rightarrow E = \left( {0 \cdot 213503\;} \right)\left( {931 \cdot 49432} \right)$
$ \Rightarrow E = 198 \cdot 87{\text{MeV}}$
$ \Rightarrow E \approx 198 \cdot 9{\text{MeV}}$
The energy released is equal to $E = 198 \cdot 9{\text{MeV}}$.

The correct answer is equal to this problem is option A.

Note:The nuclear fission is the breaking of an atom into several particles and the nuclear fusion is a process in which two or more atoms combine together to make a single particle and in this process also the energy is released but in nuclear fusion there is no byproducts which are highly radioactive like in nuclear fission. There was one neutron striking the uranium and there are three neutrons that get released therefore a total of two neutrons which take action in reaction.