In the following nuclear reactions $_7N^{14} + _2He^4 \to _8O^{17} + X_1$ and $_{13}Al^{27} + _1D^2 \to _{14}Si^{28} + X_2$,$X_1$ and $X_2$ are respectively:

$A$ closed vessel with rigid walls contains $1 \ mol$ of ${ }_{92}^{238} U$ and $1 \ mol$ of air at $298 \ K$. Considering the complete decay of ${ }_{92}^{238} U$ to ${ }_{82}^{206} Pb$, the ratio of the final pressure to the initial pressure of the system at $298 \ K$ is:

$_1H^2 + _1H^2 \to _2He^3 + _0n^1$. The above nuclear reaction is called:

What is $X$ in the following reaction?
${}_{12}Mg^{26} + {}_{1}H^{2} \longrightarrow {}_{12}Mg^{27} + X$

On bombarding $_8^{16}O$ with deuterons,the nuclei of the product formed will be

For the reaction $^{14}N + \alpha \longrightarrow {}^{17}O + p$, $1.16 \ MeV$ (Mass equivalent $= 0.00124 \ amu$) of energy is absorbed. Mass on the reactant side is $18.00567 \ amu$ and proton mass $= 1.00782 \ amu$. The atomic mass of ${}^{17}O$ will be (in $amu$)