`._(88)^(226)Ra to ._(Z)^(A)Rn+._(2)^(4)He`
Radon is prepared by `alpha`-decay from radium, in above reaction A and Z are respectively

Options

A. (A) 226, 86
B. (B) 222,90
C. (C) 222,86
D. (D) 230,90

Answer

To solve the problem of determining the values of A and Z in the alpha decay reaction of radium to radon and helium, we can follow these steps:

### Step 1: Write down the decay equation
The decay equation provided is:
\[ _{88}^{226}\text{Ra} \rightarrow _{Z}^{A}\text{Rn} + _{2}^{4}\text{He} \]

### Step 2: Identify the components
- **Radium (Ra)** has an atomic number (Z) of 88 and a mass number (A) of 226.
- **Helium (He)**, which is the alpha particle, has an atomic number of 2 and a mass number of 4.

### Step 3: Set up equations for atomic and mass numbers
From the decay process, we know:
1. The total atomic number before decay = Atomic number of Ra = 88
2. The total atomic number after decay = Atomic number of Rn (Z) + Atomic number of He (2)

This gives us the equation:
\[ 88 = Z + 2 \]

3. The total mass number before decay = Mass number of Ra = 226
4. The total mass number after decay = Mass number of Rn (A) + Mass number of He (4)

This gives us the equation:
\[ 226 = A + 4 \]

### Step 4: Solve for Z
From the first equation:
\[ Z = 88 - 2 = 86 \]

### Step 5: Solve for A
From the second equation:
\[ A = 226 - 4 = 222 \]

### Step 6: Conclusion
Thus, the values of A and Z are:
- A = 222
- Z = 86

### Step 7: Identify the correct option
Looking at the options provided:
- Option A: 226, 86
- Option B: 222, 90
- Option C: 222, 86
- Option D: 230, 90

The correct answer is **Option C: 222, 86**.

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