When Uranium is bombarded with neutrons,it undergoes fission. The fission reaction can be written as ${}_{92}U^{235} + {}_0n^1 \to {}_{56}Ba^{141} + {}_{36}Kr^{92} + 3x + Q$,where three particles named $x$ are produced and energy $Q$ is released. What is the name of the particle $x$?

When a uranium isotope ${ }_{92}^{235} U$ is bombarded with a neutron,it generates ${ }_{36}^{89} Kr$,three neutrons and:

The energy released if hydrogen atoms are combined to form $^{4}_{2}He$ is . . . . . . MeV. (Take binding energies per nucleon of $^{2}H$ and $^{4}_{2}He$ as $1.1$ MeV and $7.2$ MeV,respectively)

The energy released in the fusion of $2 \ kg$ of hydrogen deep in the sun is $E_{H}$ and the energy released in the fission of $2 \ kg$ of ${ }^{235} U$ is $E_U$. The ratio $\frac{E_H}{E_U}$ is approximately :
(Consider the fusion reaction as $4{ }_1^1 H + 2 e^{-} \rightarrow { }_2^4 He + 2 \nu + 6 \gamma + 26.7 \ MeV$,energy released in the fission reaction of ${ }^{235} U$ is $200 \ MeV$ per fission nucleus and $N_{A} = 6.023 \times 10^{23}$ )

What will be the energy released in joule,in the process of fission by $1 \text{ mg}$ of ${ }_{92}^{240} U$? Assume energy release per fission is $200 \text{ MeV}$. [Use Avogadro's number as $6 \times 10^{23}$ and $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$]

When two deuteron nuclei fuse to form a helium nucleus,energy is released because the mass of the helium nucleus is .....