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

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Group Displacement Law

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

74

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76

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78

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82

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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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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 78Decrease 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.