A nucleus with `Z = 92` emits the following in a sequence `a, beta^(bar) , beta^(bar)a,a,a,a,a, beta^(bar) , beta^(bar) , a, beta^(+) , beta^(+) , a ` Them `Z` of the resulting nucleus is
A
`76`
B
`78`
C
`82`
D
`74`
Text Solution
Verified by Experts
The correct Answer is:
B
The number of `a - particle released = 8` Therefore the atomic number should decreases by `16` The number of `beta^(bar) - particle released = 4 ` Therefore the atomic number should increases by `4` Also the number of `beta ^(+)` particle released is `2` which should decreases the atomic number is `92 - 16 + 4 - 2 = 78`
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