Two nuclei of mass number $3$ combine with another nucleus of mass number $4$ to yield a nucleus of mass number $10$. If the binding energy per nucleon for the mass numbers $3$, $4$, and $10$ are $5.6 \text{ MeV}$, $7.4 \text{ MeV}$, and $6.1 \text{ MeV}$, respectively, then in the process, $\Delta Mc^2 = . . . . . . \text{ MeV}$.

The figure shows a plot of binding energy per nucleon $E_b$ against the nuclear mass number $A$. $A, B, C, D, E, F$ correspond to different nuclei. Consider four reactions:
$(i) A + B \to C + \varepsilon$
$(ii) C \to A + B + \varepsilon$
$(iii) D + E \to F + \varepsilon$
$(iv) F \to D + E + \varepsilon$
Here, $\varepsilon$ is the energy released. In which reactions is $\varepsilon$ > 0?

The correct statement of the following is:

If the binding energy of $N^{14}$ is $7.5 \text{ MeV}$ per nucleon and that of $N^{15}$ is $7.7 \text{ MeV}$ per nucleon, then the energy required to remove a neutron from $N^{15}$ is (in $\text{ MeV}$)

If the binding energy per nucleon for $_3{Li}^7$ and $_2{He}^4$ nuclei are $5.60 \, MeV$ and $7.06 \, MeV$ respectively,then the energy released in the reaction $_3{Li}^7 + _1H^1 \rightarrow 2 \, _2{He}^4$ is ...... $MeV$.

The binding energy per nucleon of ${}_3^7Li$ and ${}_2^4He$ nuclei are $5.60\,MeV$ and $7.06\,MeV$ respectively. In the nuclear reaction ${}_3^7Li + {}_1^1H \to 2{}_2^4He + Q$, the value of energy $Q$ released is.........$MeV$.