The atomic mass of ${{}_{6}C}^{12}$ is $12.000000$ a.m.u. and that of ${{}_{6}C}^{13}$ is $13.003354$ a.m.u. The energy required (in $\mathrm{MeV}$) to remove a neutron from ${{}_{6}C}^{13}$ is (The mass of neutron is $1.008665$ a.m.u.)

A nucleus with $Z=92$ emits the following in a sequence:
$\alpha \text{,} \alpha \text{,} {\beta }^{-}\text{,} {\beta }^{-}\text{,} \alpha \text{,} \alpha \text{,} \alpha \text{,} \alpha \text{,} {\beta }^{-}\text{,} {\beta }^{-}\text{,} \alpha \text{,} {\beta }^{+}\text{,} {\beta }^{+}\text{,} \alpha$. The $Z$ of the resulting nucleus is

Binding energy per nucleon v/s mass number curve for nuclei is shown in the figure. $\mathrm{W}\text{,} \mathrm{X}$, $\mathrm{Y}$ and $\mathrm{Z}$  are four nuclei indicated on the curve. The process that would release energy is                                                                                                                                                           

A beta particle is emitted in the following decay scheme

${}^{12}\mathrm{N}\to {{}^{12}\mathrm{C}}^{*}+{\mathrm{e}}^{+}+\mathrm{v}$

${{}^{12}\mathrm{C}}^{*}\to {}^{12}\mathrm{C}+\gamma \left(4.43 \mathrm{MeV}\right)$

If the atomic mass of ${}^{12}\mathrm{N}$ is $12.018613 \mathrm{u}$, then the maximum kinetic energy of beta particle emitted, in ${}^{12}\mathrm{N}\to {}^{12}\mathrm{C}+{\mathrm{e}}^{+}+\mathrm{v}+\gamma \left(4.43 \mathrm{MeV}\right)$ is close to $\left[{\mathrm{m}}_{\mathrm{e}}=0.511\frac{\mathrm{MeV}}{{\mathrm{c}}^2}\right]$