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: In every nuclear reaction the total mass number (top numbers) and the total atomic number (bottom numbers) are conserved on both sides.

Step 2: Balance the mass numbers. Left side: \( 4 + A \). Right side: \( (A+3) + A_W \). So \[ 4 + A = (A+3) + A_W \ \Rightarrow\ A_W = 4 + A - A - 3 = 1. \]

Step 3: Balance the atomic numbers. Left side: \( 2 + Z \). Right side: \( (Z+2) + Z_W \). So \[ 2 + Z = (Z+2) + Z_W \ \Rightarrow\ Z_W = 0. \]

Step 4: Particle W has mass number 1 and charge 0, i.e. \( {}_{0}^{1}n \), which is a neutron. Hence option 4. A proton would have charge \( +1 \), an electron/positron mass number 0, so they are ruled out.

\[\boxed{W = {}_{0}^{1}n\ (\text{neutron})}\]