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

Answer

`8alpha and 6 beta`

Explanation

Let x be the number of alpha particles and y be the no. of `beta` particles emitted in the disintegration process of `._(92)U^(238)` to `._(82)Pb^(206)`. We may write
`._(92)U^(238) to._(82)Pb^(206)+(._(2)He^(4))+y(._(-1)e^(0))`
form law of conservation of mass number
`238=206+4x+y(0)=206+4x`
`=4x=238-206=32, x=8`
form law of conservation of mass number,
`92=82+2x-y=-10+2x`
`y=-10+2(8)=6`
Hence `8alpha` particles and `6 beta` particles are emitted.