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Question: In nuclear reaction \(2He^{4} +_{z}X^{A} \rightarrow_{z + 2}Y^{A + 3} + A,\mspace{6mu} A\)denotes....

In nuclear reaction 2He4+zXA→z+2YA+3+A, 6muA2He^{4} +_{z}X^{A} \rightarrow_{z + 2}Y^{A + 3} + A,\mspace{6mu} Adenotes.

A

Electron

B

Positron

C

Proton

D

Neutron

Answer

Neutron

Explanation

Solution

The given equation is 2He4+zXA→z+2YA+3+A2He^{4} +_{z}X^{A} \rightarrow_{z + 2}Y^{A + 3} + A

Applying charge and mass conservation

4+A=A+3+x⇒x=14 + A = A + 3 + x \Rightarrow x = 1⇒ 2+z=z+2+n⇒n=02 + z = z + 2 + n \Rightarrow n = 0

Hence A is a neutron.