निम्नलिखित के लिए नाभिकीय समीकरण लिखिए :

$(i)$ ${ }_{88}^{226} Ra$ का $\alpha$ -क्षय

$(ii)$ ${ }_{94}^{242} Pu$ का $\alpha$ -क्षय

$(iii)$ $\frac{32}{15} P$ का $\beta^{-}$ -क्षय

$(iv)$ $_{83}^{210} Bi$ का $\beta^{-}$ -क्षय

$(v)$ $_{6}^{11} C$ का $\beta^{+}$ -क्षय

$(vi)$ ${ }_{43}^{97} Te$ का $\beta^{+}$ -क्षय

$(vii)$ ${ }_{54}^{120} Xe$ का इलेक्ट्रॉन अभिग्रहण

$\alpha$ is a nucleus of helium $\left(_{2} H e^{4}\right)$ and $\beta$ is an electron $(e^{-}$ for $\beta^{-}$ and $e^{+}$ for $\beta^{+}$) In every $\alpha$ decay, there is a loss of $2$ protons and $2$ neutrons. In every $\beta^{+}$ -decay, there is a loss of $1$ proton and a neutrino is emitted from the nucleus. In every $\beta^{-}$ -decay, there is a gain of $1$ proton and an antineutrino is emitted from the nucleus. For the given cases, the various nuclear reactions can be written as:

$(i)\;_{88} R a^{226} \rightarrow_{86} R n^{222}+_{2} H e^{4}$

$(i i) \;^{242}_{94} Pu \rightarrow_{92}^{238} U+_{2}^{4} H e$

$(i i i)\;_{15}^{32} P \rightarrow_{16}^{32} S+e^{-}+\bar{v}$

$(i v)\;_{83}^{210} B \rightarrow_{84}^{210} P O+e^{-}+\bar{v}$

$(v)\;_{6}^{11} C \rightarrow_{5}^{11} B+e^{+}+v$

$(v i)\;_{43}^{97} T c \rightarrow_{42}^{97} M O+e^{+}+v$

$(v i i)\;_{54}^{120} X e+e^{+} \rightarrow_{53}^{120} I+v$