The binding energy $(BE)$ per nucleon for an element is $7.14 \text{ MeV}$. If the total $BE$ of the element is $28.6 \text{ MeV}$, then the number of nucleons in the element is:

$A$ plot of the number of neutrons $(N)$ against the number of protons $(Z)$ for stable nuclei exhibits upward deviation from linearity for atomic number $Z > 20$. For an unstable nucleus having an $N/Z$ ratio less than $1$,the possible mode$(s)$ of decay is(are):
$(A)$ $\beta^{-}$-decay ($\beta$ emission)
$(B)$ Orbital or $K$-electron capture
$(C)$ Neutron emission
$(D)$ $\beta^{+}$-decay (positron emission)

In the nuclear reaction,$_1H^2 + _1H^2 \to _0n^1 + _2He^3$. If the binding energy of deuteron is $2.23 \ MeV$ and the $Q$-value of the reaction is $3.27 \ MeV$,then the binding energy of $_2He^3$ is ......... $MeV$.

The mass defect of $ { }_{2}^{4} He $ is $ 0.03 \ u $. The binding energy per nucleon of helium (in $ MeV $ ) is

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range $30$ to $170$.
Reason $R:$ Nuclear force is short-ranged.
In the light of the above statements, choose the correct answer from the options given below:

According to the mass-energy equivalence relation,$9 \times 10^{13} \text{ J}$ of energy can be converted into $\qquad$ maximum mass. [Speed of light $c = 3 \times 10^{8} \text{ m/s}$] (in $\text{ g}$)