A radioactive element `._90 X^238` decay into `._83 Y^222`. The number of `beta-`particles emitted are.

Options

A. 1
B. 2
C. 4
D. 6

Answer

To solve the problem of determining the number of beta particles emitted during the decay of the radioactive element \( _{90}^{238}X \) into \( _{83}^{222}Y \), we can follow these steps:

### Step 1: Understand the decay process
The decay of a radioactive element can involve the emission of alpha particles, beta particles, or gamma rays. In this case, we need to analyze the changes in both the atomic number and the mass number of the element \( X \) as it transforms into element \( Y \).

### Step 2: Write down the initial and final states
- Initial state: \( _{90}^{238}X \)
- Final state: \( _{83}^{222}Y \)

### Step 3: Calculate the change in mass number
The mass number of element \( X \) is 238, and the mass number of element \( Y \) is 222. We can find the change in mass number:
\[ \text{Change in mass number} = 238 - 222 = 16 \]

### Step 4: Determine the number of alpha particles emitted
Each alpha particle emission decreases the mass number by 4. To find the number of alpha particles emitted, we divide the total change in mass number by the change caused by one alpha particle:
\[ \text{Number of alpha particles} = \frac{16}{4} = 4 \]

### Step 5: Calculate the change in atomic number
The atomic number of element \( X \) is 90. Since 4 alpha particles are emitted, each causing a decrease of 2 in the atomic number, we calculate the total decrease:
\[ \text{Change in atomic number} = 4 \times 2 = 8 \]
Thus, the new atomic number after the emission of 4 alpha particles is:
\[ 90 - 8 = 82 \]

### Step 6: Determine the atomic number of the intermediate element
After the emission of 4 alpha particles, we have an intermediate element \( Z \) with:
- Atomic number = 82
- Mass number = 222

### Step 7: Transition from intermediate element \( Z \) to \( Y \)
Now, we need to transition from \( Z \) (which has atomic number 82) to \( Y \) (which has atomic number 83). This transition requires the emission of beta particles. Each beta particle increases the atomic number by 1.

### Step 8: Calculate the number of beta particles emitted
To go from atomic number 82 to atomic number 83, we need:
\[ \text{Number of beta particles} = 83 - 82 = 1 \]

### Final Answer
Thus, the total number of beta particles emitted during the decay of \( _{90}^{238}X \) into \( _{83}^{222}Y \) is **1**.

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