Question

An atom of radon gas emits an alpha particle as shown below. Radon-222 has 86 protons and 136 neutrons. What daughter product will form from this decay? Francium-222 (87 protons) Polonium-218 (84 protons) Radium-226 (88 protons) Lead-220 (82 protons)

Answer 1

Answer is Polonium-218 (84 protons).

alpha decay is an emission of ₂⁴He nucleus. If an element undergoes an alpha decay, the mass of the daughter nucleus formed is reduced by 4 compared to mass of parent atom and atomic number is reduced by 2 compared to atomic number of parent atom.

Since the parent atom has 86 protons, atomic number is 86. Hence, the daughter atom should have 86 - 2 = 84 as atomic number. The parent atom has 222 as its mass. Hence, formed daughter atom should have 222 - 4 = 218 as its mass number.

The reaction for this decay is;

₈₆²²²Rn → ₈₄²¹⁸Po + ₂⁴∝

Answer 2

The answer is Polonium-218 (84 protons).

Alpha particles consist of two protons and two neutrons. So, mass is 4 (2+2) amu $_{2}^{4} \alpha$.
When an atom of emits an alpha particle, its atomic number will be reduced by 2 and a mass number will be reduced by 4:
$_{Z} ^{A} X$ → $_{Z-2} ^{Y-4} Y$ + $_{2}^{4} \alpha$.

After emission of the alpha particle from Radon-222 with 86 protons, the daughter product will have the atomic number reduced by 2: 86-2 = 84 and the mass number reduced by 4: 222-4=218:
 $_{86} ^{222} Rn$ → $_{84} ^{218} Po$ + $_{2}^{4} \alpha$.