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In the given radioactive disintegration ...

In the given radioactive disintegration series,
`._90^(232)Th to _(2)^(208)Pb`
Calculate value of `(n+2)`.
Where value of n is number of isobars formed in this series, suppose there is successive emission of `beta-`particles.

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The correct Answer is:
`6alpha,4 beta`
`._(0)Th^(232)rarr` `._(82)Pb^(208)+x_(2)He^(4)+y` `._(-1)e^(0)`
`232=208=4x`
`90=82+2x-y`
Solve for `x` and `y`, we get
`x=6, y=4 `
`:. 6 alpha` and `4 beta`.
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