On fission,a $U^{235}$ nucleus releases $3 \times 10^{-11} \, J$ of energy. In a $1 \, GW$ nuclear reactor,$4.2 \%$ of this energy is converted to useful energy. The $U^{235}$ consumed (in grams) in half an hour is closest to (Avogadro number $N_A = 6.023 \times 10^{23}$)

If the binding energy per nucleon of deuteron $({ }_1 H^2)$ is $1.15 \text{ MeV}$ and an $\alpha$-particle has a binding energy of $7.1 \text{ MeV}$ per nucleon,then the energy released per nucleon in the given reaction is ${ }_1 H^2 + { }_1 H^2 \rightarrow { }_2 He^4 + Q$. (in $\text{ MeV}$)

The nucleus of $_6C^{12}$ absorbs a neutron and emits a $\beta$-particle. The resulting nucleus is ........

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

$A$ heavy nucleus $N$,at rest,undergoes fission $N \rightarrow P+Q$,where $P$ and $Q$ are two lighter nuclei. Let $\delta=M_N-M_P-M_Q$,where $M_P, M_Q$ and $M_N$ are the masses of $P, Q$ and $N$,respectively. $E_P$ and $E_Q$ are the kinetic energies of $P$ and $Q$,respectively. The speeds of $P$ and $Q$ are $v_P$ and $v_Q$,respectively. If $c$ is the speed of light,which of the following statement$(s)$ is(are) correct?
$(A)$ $E_P+E_Q=c^2 \delta$
$(B)$ $E_P=\left(\frac{M_P}{M_P+M_Q}\right) c^2 \delta$
$(C)$ $\frac{v_P}{v_Q}=\frac{M_Q}{M_P}$
$(D)$ The magnitude of momentum for $P$ as well as $Q$ is $c \sqrt{2 \mu \delta}$,where $\mu=\frac{M_P M_Q}{M_P+M_Q}$

$A$ nuclear reactor generates power by fission of ${}_{92}U^{235}$ into two equal fragments of ${}_{46}Pd^{116}$ with the emission of two $\gamma$-rays of $5.2 \text{ MeV}$ each and three neutrons. The average binding energies per nucleon of ${}_{92}U^{235}$ and ${}_{46}Pd^{116}$ are $7.2 \text{ MeV}$ and $8.2 \text{ MeV}$ respectively. The usable energy released per fission event is ......... $\text{MeV}$.