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A nucleus with Z = 92 emits the followin...

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 `
Then `Z` of the resulting nucleus is

A

74

B

76

C

78

D

82

Text Solution

Verified by Experts

The correct Answer is:
C
( c) `Z_("Resulting nucleus") = 92 - 8 xx 2 + 4 xx 1 - 2 xx 1 = 78`.
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A2Z-NUCLEAR PHYSICS-Radioactivity
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