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`._(90)Th^(232) to ._(82)Pb^(208)`. The number of `alpha and beta-"particles"` emitted during the above reaction isA. `8alpha and 4beta`B. `8alpha and 16beta`C. `4alpha and 2beta`D. `6alpha and 4beta` |
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Answer» 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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