In a radioactive series, `._(92)U^(238)` charges to `._(82)Pb^(206)` through `n_(1)`(`alpha`-decay processes) and `n_(2)`(`beta-`deacy processes)

Question

In a radioactive series, `._(92)U^(238)` charges to `._(82)Pb^(206)` through `n_(1)`(`alpha`-decay processes) and `n_(2)`(`beta-`deacy processes)

In a radioactive series, `._(92)U^(238)` charges to `._(82)Pb^(206)` through `n_(1)`(`alpha`-decay processes) and `n_(2)`(`beta-`deacy processes)
A. `n_(1) = 8,n_(2) = 8`
B. `n_(1) =6, n_(2) = 2`
C. `n_(1) = 8,n_(2) = 6`
D. `n_(1) = 6,n_(2) =8`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
`._(92)U^(238)rarr._(82)Pb^(206)`
`(238-206)/(4) = 8 alpha-`particles
Decrease in `Z` due to `alpha`-particle `= 16` but decrease in only `92-82 = 10` i.e `6beta-`particles
`n_(1) = 8, n_(2) = 6`

4 views Answered 2023-02-05 15:29