Question

In the given nuclear reaction, particle W is: \[ {}_{2}^{4}\text{He} + {}_{Z}^{A}X \ \rightarrow\ {}_{Z+2}^{A+3}Y + W \]

• A. Electron
• B. Proton
• C. Positron
• D. Neutron

Hint:

Conserve mass number and atomic number separately; the missing particle has mass number 1 and charge 0.

Answer 1

The Correct Option is D

Step 1: Write W as \({}_{z_W}^{a_W}W\) and demand that the subscripts (charge) and superscripts (mass number) match across the arrow.

Step 2: Superscript equation: incoming \(4 + A\) must equal outgoing \((A+3) + a_W\). Cancelling \(A\) gives \(a_W = 1\).

Step 3: Subscript equation: incoming \(2 + Z\) must equal outgoing \((Z+2) + z_W\). Cancelling \(Z\) gives \(z_W = 0\).

Step 4: A particle carrying one unit of mass and zero charge is the neutron \({}_{0}^{1}n\). Therefore W is a neutron.

\[\boxed{W = \text{neutron}}\]