$^{226}_{88}Ra$ → $^x_{86}Rn$ + alpha particle

Analysis

This is a nuclear chemistry problem involving radioactive decay. We are given the decay of Radium ($Ra$) into Radon ($Rn$) and an alpha particle. We need to find the value of $x$, which represents the mass number of the Radon isotope produced.

Concept

In any nuclear reaction, two fundamental laws must be satisfied:

1. Conservation of Mass Number: The sum of the mass numbers (the top numbers) on the reactant side must equal the sum of the mass numbers on the product side.
2. Conservation of Atomic Number: The sum of the atomic numbers (the bottom numbers) on the reactant side must equal the sum of the atomic numbers on the product side.

An alpha particle is a helium nucleus, represented as $^4_2He$.

Solution

The equation is: $^{226}_{88}Ra \rightarrow ^x_{86}Rn + ^4_2He$

Step 1: Set up the equation for the mass numbers (top numbers). The total mass number on the left is 226. The total mass number on the right is $x + 4$.

Equating them: $226 = x + 4$

Step 2: Solve for $x$. $x = 226 - 4$ $x = 222$

Step 3: Verify with atomic numbers (bottom numbers). The total atomic number on the left is 88. The total atomic number on the right is $86 + 2 = 88$. Since $88 = 88$, the equation is balanced.

The value of $x$ is 222.

Answer

Comparing our result to the given options: A. 226 B. 220 C. 227 D. 222

The correct option is D.

Exam Tip

Whenever you see "alpha decay," always remember to subtract 4 from the mass number and 2 from the atomic number of the parent nucleus. If it were "beta decay," the mass number remains the same, but the atomic number increases by 1!