Complete the following nuclear reaction for $\alpha$ and $\beta$ decay:
(i) ${}_{92}^{238}U⟶?+{}_2^{4}He+Q$
(ii) ${}_{11}^{22}Na⟶{}_{10}^{22}Ne+?+v$

i)  ${}_{92}^{238}U⟶?{+}_2^{4}He+\lambda$

Since ${}^{\prime }{U}^{\prime }$ gines Helium ${+}^{\prime }{Q}^{\prime }$

$\therefore$ It gives:-

${}_{92}^{238}U{⟶}_{90}^{234}Th{+}_2^{4}He+Q$

 

ii)  ${}_{11}^{22}Na{⟶}_{10}^{22}Ne+?+V$

$\beta$ Particle will be given.

${}_{11}^{22}Na{⟶}_{11}^{22}Ne+{\beta }^{+}+V$