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By the successive disintegration of ""(9...

By the successive disintegration of `""_(92)U^(238),` the final product obtained is `""_(82)Pb^(206),` then how many number of `alpha and beta`-particles are emitted?

Text Solution

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`._(92)U^(238)rarr ._(82)Pb^(214)=x._(2)He^(4)+y._(-1)e^(0)`
Equating atomic number on both sides ltbr. `238=214=xxx4+yxx0`
`:. X=6`
Hence, `6 alpha-` particles are emitted.
Equating atomic number on both sides ltbr. `92=82+6xx2+y(-1)`
`:. y=2`
Hence, `2 beta-` particles are emitted.
Therefore, total 8 particles `(6alpha,2 beta)` are emitted.
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