After losing a number of `alpha and brta`-particles, `U_92^238` changed to `Pb_32^206`. 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 determining the total number of alpha and beta particles lost when Uranium-238 decays to Lead-206, we can follow these steps:

### Step 1: Write the decay equation
The decay of Uranium-238 (U-238) to Lead-206 (Pb-206) can be represented as:
\[ \text{U}_{92}^{238} \rightarrow \text{Pb}_{82}^{206} \]

### Step 2: Determine the change in mass number
The mass number of Uranium-238 is 238, and that of Lead-206 is 206. The change in mass number is:
\[ \Delta A = 238 - 206 = 32 \]

### Step 3: Calculate the number of alpha particles emitted
Each alpha particle (α) emitted decreases the mass number by 4. If we let \( x \) be the number of alpha particles emitted, the equation for the mass number becomes:
\[ 238 - 4x = 206 \]
Solving for \( x \):
\[ 4x = 238 - 206 \]
\[ 4x = 32 \]
\[ x = \frac{32}{4} = 8 \]
Thus, 8 alpha particles are emitted.

### Step 4: Determine the change in atomic number
The atomic number of Uranium-238 is 92, and that of Lead-206 is 82. The change in atomic number is:
\[ \Delta Z = 92 - 82 = 10 \]

### Step 5: Calculate the number of beta particles emitted
Each beta particle (β) emitted increases the atomic number by 1. If we let \( y \) be the number of beta particles emitted, the equation for the atomic number becomes:
\[ 92 - 2x + y = 82 \]
Substituting \( x = 8 \):
\[ 92 - 2(8) + y = 82 \]
\[ 92 - 16 + y = 82 \]
\[ 76 + y = 82 \]
\[ y = 82 - 76 = 6 \]
Thus, 6 beta particles are emitted.

### Step 6: Calculate the total number of particles lost
The total number of particles lost is the sum of the alpha and beta particles:
\[ \text{Total particles} = x + y = 8 + 6 = 14 \]

### Final Answer
The total number of particles lost in this process is **14**.

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