${U^{238}}$ decays into $T{h^{234}}$ by the emission of an $\alpha - $ particle. There follows a chain of further radioactive decays, either by $\alpha - $ decay or by $\beta $ - decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. Which of the following stable nuclides is the and product of the ${U^{238}}$ radioactive decay chain

Assertion : Radioactive nuclei emit ${\beta ^ - }$ particles.

Reason : Electrons exist inside the nucleus

$^{22}Ne$ nucleus after absorbing energy decays into two $\alpha - $ particles and an unknown nucleus. The unknown nucleus is

${ }_{92} U^{235}$ nucleus absorbs a neutron and disintegrates into ${ }_{54} X e^{139} ,{ }_{38} Sr ^{94}$ and $X$. What will be the product $X$ ?

Which can pass through $20 \,cm$ thickness of the steel

In the nuclear decay given below 

$_z{X^A}{ \to _{z + 1}}{Y^A}{ \to _{z - 1}}{K^{A - 4}}{ \to _{z - 1}}{K^{A - 4}}$

the particles emitted in the sequence are