The binding energy per nucleon for a deuteron $(_{1}^{2}H)$ and an $\alpha -$ particle $(_{2}^{4}He)$ are $x_1$ and $x_2$ respectively. The energy $(Q)$ released in the reaction $_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{4}He + Q$ is

One microgram of matter converted into energy will give:

Which element has the maximum binding energy per nucleon? Write down its value.

The graph of binding energy per nucleon $B_N$ versus mass number $A$ is given. In which process will energy be released?

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 found to be practically independent of the atomic number $A$,for nuclei with mass numbers between $30$ and $170$.
Reason $(R) :$ Nuclear force is long range.
In the light of the above statements,choose the correct answer from the options given below $:$

For a nucleus ${ }_Z^A X$ having mass number $A$ and atomic number $Z$:
$A.$ The surface energy per nucleon $(b_s) = a_1 A^{2/3}$
$B.$ The Coulomb contribution to the binding energy $b_c = -a_2 \frac{Z(Z-1)}{A^{4/3}}$
$C.$ The volume energy $b_v = a_3 A$
$D.$ Decrease in the binding energy is proportional to surface area.
$E.$ While estimating the surface energy,it is assumed that each nucleon interacts with $12$ nucleons,($a_1, a_2$ and $a_3$ are constants)
Choose the most appropriate answer from the options given below: