How many `alpha`-particles are emitted in the nuclear transformation: `._(84)Po^(215) rarr ._(82)Pb^(211) + ? ._(2)He^(4)`

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem of how many alpha particles are emitted in the nuclear transformation of Polonium-215 to Lead-211, we can follow these steps:

### Step 1: Write down the nuclear transformation equation.
The nuclear transformation given is:
\[ _{84}^{215}\text{Po} \rightarrow _{82}^{211}\text{Pb} + ? + _{2}^{4}\text{He} \]

### Step 2: Identify the particles involved.
In this transformation:
- Polonium (Po) has an atomic number of 84 and a mass number of 215.
- Lead (Pb) has an atomic number of 82 and a mass number of 211.
- The emitted particle is an alpha particle, which is represented as Helium-4 (\(_{2}^{4}\text{He}\)).

### Step 3: Analyze the changes in atomic and mass numbers.
When an alpha particle is emitted:
- The atomic number decreases by 2 (because an alpha particle has 2 protons).
- The mass number decreases by 4 (because an alpha particle has a mass of 4).

### Step 4: Set up the equations for atomic and mass numbers.
**For atomic numbers:**
Initial atomic number (Po) = 84
Final atomic number (Pb) + Atomic number of alpha particle = 82 + 2
\[ 84 = 82 + 2 \]
This equation is balanced.

**For mass numbers:**
Initial mass number (Po) = 215
Final mass number (Pb) + Mass number of alpha particle = 211 + 4
\[ 215 = 211 + 4 \]
This equation is also balanced.

### Step 5: Determine the number of alpha particles emitted.
Since both atomic and mass numbers are balanced with the emission of one alpha particle, we conclude that:
- Only **one alpha particle** is emitted in this transformation.

### Final Answer:
The number of alpha particles emitted is **1**.

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