In the given nuclear reaction, the element X is Na 11 22 → X + e + + v

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Question: Answered & Verified by Expert

In the given nuclear reaction, the element

X

is

{}_{11}^{22}{Na} \to X + e^{+} + v

PhysicsNuclear PhysicsNEETNEET 2022 (Phase 1)

Options:

A ${}_{10}^{23}{Ne}$
B ${}_{10}^{22}{Ne}$
C ${}_{12}^{22}{Mg}$
D ${}_{11}^{23}{Na}$

Solution:

1109 Upvotes Verified Answer

The correct answer is:

{}_{10}^{22}{Ne}

As positron is emitting, it is positive beta decay with equation,

${}_{z}^{A}X ⟶ {}_{z - 1}^{A}Y + e^{+} + ν$

Hence, $\Rightarrow {}_{11}^{22}{Na} ⟶ {}_{10}^{22}{Ne} + e^{+} + ν$

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