In the nuclear reaction $_{92}U^{238} \to _{82}Pb^{206} + x\,_{2}He^{4} + y\,_{-1}\beta^{0}$,the values of $x$ and $y$ are respectively......

$_{5}B^{10} + _{2}He^{4} \rightarrow X + _{0}n^{1}$
In the above nuclear reaction '$X$' will be

The radioactive isotope $_{27}^{60}Co$,which is used in the treatment of cancer,can be produced by an $(n, p)$ reaction. For this reaction,the target nucleus is:

$_6C^{14}$ is formed from $_7N^{14}$ in the upper atmosphere by the action of the fundamental particle:

$_{13}Al^{28}$ when radiated by a suitable projectile gives $_{15}P^{31}$ and a neutron. The projectile used is:

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 $amu$)