After losing a number of `alpha and brta`-particles, `U_92^238` changed to `Pb_82^206`. The total number of particles lost in this process is

Options

A. 14
B. 5
C. 8
D. 32

Answer

To solve the problem of how many alpha and beta particles are lost when Uranium-238 (U-238) transforms into Lead-206 (Pb-206), we can follow these steps:

### Step 1: Understand the Reaction
Uranium-238 has an atomic number of 92 and a mass number of 238. It transforms into Lead-206, which has an atomic number of 82 and a mass number of 206. We need to find out how many alpha (α) and beta (β) particles are emitted during this transformation.

### Step 2: Define Variables
Let:
- \( X \) = number of alpha particles emitted
- \( Y \) = number of beta particles emitted

### Step 3: Conservation of Mass Number
The mass number before the reaction must equal the mass number after the reaction. The mass number of alpha particles is 4, and beta particles have a mass number of 0. Therefore, we can set up the equation:

\[ 238 = 206 + 4X + 0Y \]

This simplifies to:

\[ 4X = 238 - 206 \]
\[ 4X = 32 \]
\[ X = \frac{32}{4} = 8 \]

So, the number of alpha particles emitted is \( X = 8 \).

### Step 4: Conservation of Atomic Number
Next, we apply the conservation of atomic number. The atomic number of alpha particles is 2, and the atomic number of beta particles is -1. Therefore, we can set up the equation:

\[ 92 = 82 + 2X - Y \]

Substituting \( X = 8 \):

\[ 92 = 82 + 2(8) - Y \]
\[ 92 = 82 + 16 - Y \]
\[ 92 = 98 - Y \]
\[ Y = 98 - 92 = 6 \]

So, the number of beta particles emitted is \( Y = 6 \).

### Step 5: Total Particles Lost
Now, we can find the total number of particles lost:

\[ \text{Total particles lost} = X + Y = 8 + 6 = 14 \]

### Conclusion
The total number of particles lost in the process is **14**.

---