A nucleus with Z -92 emits the following in a sequence - - - - - - - - - - - - - -The Z of the resulting nucleus isA- 74B- 76C- 78D- 82

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Beta Decay

A nucleus wit...

Question

A nucleus with Z = 92 emits the following in a sequence :
$α, β -, β -, α, α, α, α, α, β -, β -, β +, α, β +, α$
The Z of the resulting nucleus is

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Solution

The correct option is C
78

Decrease in Z due to emission of 8 $α$ - particles = 16

Increase in Z due to emission of 4 $β$ - particles = 4

Decrease in Z due to emission of 2 positrons = 2

$∴$ Z of resultant nucleus = 92 – 16 + 4 – 2 = 78

So, (c) is the right choice.

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Q. A nucleus with

Z = 92

, emits the following in a sequence,

α, β^{-}, β^{-}, α, α, α, α, α, β^{-},

β^{-}, α, β^{+}, β^{+}, α

. The

Z

of the resulting nucleus is ______.

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A nucleus with Z = 92 emits the following in a sequence : α,β−,β−,α,α,α,α,α,β−,β−,β+,α,β+,α The Z of the resulting nucleus is

The correct option is C 78 Decrease in Z due to emission of 8 α - particles = 16 Increase in Z due to emission of 4 β - particles = 4 Decrease in Z due to emission of 2 positrons = 2 ∴ Z of resultant nucleus = 92 – 16 + 4 – 2 = 78 So, (c) is the right choice.

A nucleus with Z = 92 emits the following in a sequence :
$α, β -, β -, α, α, α, α, α, β -, β -, β +, α, β +, α$
The Z of the resulting nucleus is

The correct option is C
78

Decrease in Z due to emission of 8 $α$ - particles = 16

Increase in Z due to emission of 4 $β$ - particles = 4

Decrease in Z due to emission of 2 positrons = 2

$∴$ Z of resultant nucleus = 92 – 16 + 4 – 2 = 78

So, (c) is the right choice.