$^{226}_{88}Ra$ emits two alpha particles to produce polonium(po) nuclide. The correct formula for the nuclide?

Analyze the question: Radium-226 (atomic number 88) emits two alpha particles. We need the nuclide that remains, which is said to be polonium.

Concept: An alpha particle is a helium nucleus, ⁴₂He. Each alpha emission reduces the mass number by 4 and the atomic number by 2.

Solution:

Two alpha particles are lost, so:

• Mass number: 226 − 2(4) = 226 − 8 = 218
• Atomic number: 88 − 2(2) = 88 − 4 = 84

Atomic number 84 is polonium, which matches the question.

Equation:

²²⁶₈₈Ra → ²¹⁸₈₄Po + 2 ⁴₂He

(The steps are Ra-226 → Rn-222 → Po-218.)

Checking the options:

• A (224, 86) and B (222, 86) have atomic number 86, which is radon, not polonium.
• C (222, 84) has the right atomic number but the wrong mass number. It would require only one alpha emission for mass, which is inconsistent.
• D (218, 84) matches our calculation.

Answer: D, ²¹⁸₈₄Po.

Tip: For alpha decay, subtract 4 from the mass number and 2 from the atomic number for each alpha particle. Always multiply by the number of particles emitted. Here the key is not to stop after the first alpha, which would give radon-222.