91Ó°ÊÓ

We need to find balanced nuclear equation for the alpha decay of given element${}_{84}{}^{208}Po$ .

The unbalanced equation is

$Polonium-208→alphaparticle+decayproduct$

Let the decay product be $X$

The superscript of $X$must be$208-4=204$

The subscript of $X$must be $84-2=82$.

Element 82 is Pb. So the balanced equation is

${}_{84}{}^{208}Po→{}_2{}^{4}He+{}_{82}{}^{204}Pb$

We need to find balanced nuclear equation for the alpha decay of given element ${}_{90}{}^{232}Th$.

The unbalanced equation is

$Thorium-232→alphaparticle+decayproduct$

Let the decay product be$X$

The superscript of $X$must be $232-4=228$.

The subscript of $X$must be $90-2=88$.

Element 88 is Radium. So the balanced equation is

${}_{90}{}^{232}Th→{}_{88}{}^{228}Ra+{}_2{}^{4}He$

We need to find balanced nuclear equation for the alpha decay of given element ${}_{102}{}^{251}No$.

The unbalanced equation is

$Nobilium-251→aplhaparticle+decayproduct$

Let the decay particle be$X$

The superscript of $X$must be $251-4=247$.

The subscript of X must be $102-2=100$.

Element 100 is Fm. So the equation is

${}_{102}{}^{251}No→{}_2{}^{4}He+{}_{100}{}^{247}Fm$

We need to find balanced nuclear equation for the alpha decay of given element radon-220.

The unbalanced equation is

$Radon-220→alphaparticle+decayproduct$

Let the decay product be$X$

The superscript of $X$must be $220-4=216$.

The subscript of $X$must be $86-2=84$.

Element 84 is Po. So the equation is

${}_{86}{}^{220}Rn→{}_2{}^{4}He+{}_{84}{}^{216}Po$