$^{226}_{88}Ra$ $\to$ $^{222}_{86}Rn$ + X + Energy. The equation above represents a nuclear decay of radium isotope. What is X?

Step 1: Analyze the question Radium-226 decays into radon-222 and a particle X. We need to identify X using conservation of mass number (A) and atomic number (Z).

Step 2: The concept In any nuclear reaction:

• Total mass number (top number) is conserved.
• Total atomic number (bottom number) is conserved.

Step 3: Solve

Mass number: 226 = 222 + A, so A = 4

Atomic number: 88 = 86 + Z, so Z = 2

So X has mass number 4 and atomic number 2, which is ⁴₂He. That is a helium nucleus, which is an alpha particle.

Step 4: Check the other options

• Beta particle (⁰₋₁e): mass number 0, so it would not account for the drop of 4 in mass number.
• Gamma ray and X-ray: no mass and no charge, so the numbers would not balance.
• Neutron (¹₀n): mass number 1 and atomic number 0, which does not balance either.

Answer: A. Alpha particle

Tip: If the mass number drops by 4 and the atomic number drops by 2, it is always alpha decay. If the mass number stays the same and the atomic number rises by 1, it is beta decay.