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

$A$ $^{235}U$ nuclear reactor generates energy at a rate of $3.70 \times 10^7 \text{ J/s}$. Each fission liberates $185 \text{ MeV}$ of useful energy. If the reactor has to operate for $144 \times 10^4 \text{ s}$, the mass of the fuel needed is (Assume Avogadro's number $= 6 \times 10^{23} \text{ mol}^{-1}$, $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$) (in $\text{ kg}$)

When a proton collides with a $_3Li^7$ nucleus and results in a $_4Be^8$ nucleus,the emitted particle is:

$U^{235}$ nuclear reactor generates energy at a rate of $3.70 \times 10^7 \text{ J/s}$. Each fission liberates $185 \text{ MeV}$ of useful energy. If the reactor has to operate for $144 \times 10^4 \text{ s}$, then the mass of the fuel needed is (Assume Avogadro's number $= 6 \times 10^{23} \text{ mol}^{-1}$, $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$) (in $\text{ kg}$)

The nuclear reaction $^2H + ^2H \to ^4He$ (mass of deuteron $= 2.0141 \,amu$ and mass of $He = 4.0024 \,amu$) is

Suppose a ${ }_{88}^{226} Ra$ nucleus at rest and in the ground state undergoes $\alpha$-decay to a ${ }_{86}^{222} Rn$ nucleus in its excited state. The kinetic energy of the emitted $\alpha$ particle is found to be $4.44 \text{ MeV}$. The ${ }_{86}^{222} Rn$ nucleus then goes to its ground state by $\gamma$-decay. The energy of the emitted $\gamma$-photon is . . . . . . . $\text{keV}$.
[Given: atomic mass of ${ }_{88}^{226} Ra = 226.005 \text{ u}$,atomic mass of ${ }_{86}^{222} Rn = 222.000 \text{ u}$,atomic mass of $\alpha$ particle $= 4.000 \text{ u}$,$1 \text{ u} = 931 \text{ MeV}/c^2$]