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`._(90)Th^(232) to ._(82)Pb^(208)`. The number of `alpha and beta-"particles"` emitted during the above reaction is

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`._(90)Th^(232) to ._(82)Pb^(208)`. The number of `alpha and beta-"particles"` emitted during the above reaction is
A. `8alpha and 4beta`
B. `8alpha and 16beta`
C. `4alpha and 2beta`
D. `6alpha and 4beta`
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Correct Answer - D
No. of `alpha`-particle ` =(232-208)/(4)=(24)/(4)=6alpha`
No. of `beta`-particle `= `2xx6-(90-82)`
`=12-8`
=`4beta`
`"Hence", 6alpha and 4 beta` particle are emitted.
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