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

An element $_{96}X^{227}$ emits $4\alpha$ and $5\beta$ particles to form a new element $Y$. The atomic number and mass number of $Y$ are:

If $_{92}U^{235}$ nucleus absorbs a neutron and disintegrates into $_{54}Xe^{139}$,$_{38}Sr^{94}$ and $X$,then what will be the product $X$?

Deuterons when bombarded on a nuclide produce $_{18}Ar^{38}$ and neutrons. The target is

The number of $\alpha$ and $\beta$-particles emitted during the transformation of $_{90}Th^{232}$ to $_{82}Pb^{208}$ are respectively

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 $amu$)