In the final Uranium radioactive series the initial nucleus is `U_(92)^(238) ` and the final nucleus is `Pb_(82)^(206)` . When Uranium nucleus decays to lead , the number of a - particle is …….. And the number of `beta` - particles emitted is ……

Explanation

Let a and b be the number of `alpha-` and `beta-` particels emitted when `._(92)U^(238)` decays to `._(82)Pb^(206)`. We know that
(a) The emission of an `alpha-` particle `(._2He^(4))` decreases the charege number by two and mass number by four. Therefore,emission of a `alpha-particles` will reduce the charge number by `2`a and mass number by `4a`.
(b) The emission of a `beta-` particle increases the charege number by one and leaves the mass number unchanged. Therefore,emission of a `beta`-particles will increases the charge number by `b xx 1=b`.
Thus, `._(92)U^(238) rarr ._(82)Pb^(206) +a(._(2)He^(4))=b(._(-1) beta^(0))`
Applying the law of conservation of charege number and mass number before and after the decay, we have
`92=82+2a -b`
`238 =206 + 4a`
From Eq. (ii),`a=8`,i.e., number of emitted `alpha` - particles `=8`
From Eq. (i),`b=6`,i.e., number of emitted `beta` - particles `=6`.