The number of `alpha- and beta-`particles emitted when a radioactive element `._(90)E^(322)` changes into `._(86)G^(220)` will be

Options

A. 5 and 4
B. 2 and 3
C. 3 and 2
D. 4 and 1

Explanation

To determine the number of alpha and beta particles emitted when the radioactive element \(_{90}E^{322}\) changes into \(_{86}G^{220}\), we can follow these steps:

### Step 1: Identify the Initial and Final Nuclei
- The initial nucleus is \(_{90}E^{322}\).
- The final nucleus is \(_{86}G^{220}\).

### Step 2: Write the Nuclear Reaction
The general form of the decay can be written as:
\[ _{90}E^{322} \rightarrow _{86}G^{220} + \text{(alpha and beta particles)} \]

### Step 3: Calculate the Change in Mass Number and Atomic Number
- The mass number of the initial nucleus is 322, and that of the final nucleus is 220.
- The change in mass number (A) is:
\[ \Delta A = 322 - 220 = 102 \]
- The atomic number of the initial nucleus is 90, and that of the final nucleus is 86.
- The change in atomic number (Z) is:
\[ \Delta Z = 90 - 86 = 4 \]

### Step 4: Determine the Number of Alpha Particles Emitted
Each alpha particle (\(_{2}He^{4}\)) emitted decreases the mass number by 4 and the atomic number by 2. Let \(x\) be the number of alpha particles emitted. The equations for the changes can be set up as follows:
\[ 4x = 102 \quad \text{(for mass number)} \]
\[ 2x = 4 \quad \text{(for atomic number)} \]

From the second equation:
\[ x = 2 \]
Thus, 2 alpha particles are emitted.

### Step 5: Determine the Number of Beta Particles Emitted
After emitting 2 alpha particles, the new atomic number becomes:
\[ 90 - 2 \times 2 = 86 \]
Now, since we need to maintain the mass number at 220, we can calculate the remaining change needed. Each beta particle (\(_{-1}e^{0}\)) emitted increases the atomic number by 1 without changing the mass number. Let \(y\) be the number of beta particles emitted. The equation for the atomic number becomes:
\[ 86 + y = 86 \]
This implies:
\[ y = 0 \]

### Conclusion
Thus, the total emissions are:
- Alpha particles emitted: 2
- Beta particles emitted: 0

### Final Answer
The number of alpha and beta particles emitted is:
- Alpha particles: 2
- Beta particles: 0

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