The isotope `""_(92)^(235) U` decays in a number of steps to an isotope of `""_(87)^(207)Pb`. The no of particles emitted in this process will be :

Options

A. `4 alpha`
B. `6 beta`
C. `7 alpha`
D. `4 beta`

Explanation

To determine the number of particles emitted during the decay of the isotope \( _{92}^{235}U \) to \( _{82}^{207}Pb \), we need to analyze the decay process in terms of alpha and beta emissions.

### Step-by-Step Solution:

1. **Identify the Initial and Final Isotopes**:
- Initial isotope: \( _{92}^{235}U \) (Uranium-235)
- Final isotope: \( _{82}^{207}Pb \) (Lead-207)

2. **Determine the Change in Mass Number**:
- The mass number of Uranium-235 is 235, and that of Lead-207 is 207.
- The change in mass number = Initial mass number - Final mass number
\[ \Delta A = 235 - 207 = 28 \]

3. **Calculate the Number of Alpha Particles Emitted**:
- Each alpha particle (\( \alpha \)) has a mass number of 4.
- Let \( a \) be the number of alpha particles emitted.
- The equation for the mass number becomes:
\[ 235 - 4a = 207 \]
- Rearranging gives:
\[ 4a = 235 - 207 = 28 \implies a = \frac{28}{4} = 7 \]

4. **Determine the Change in Atomic Number**:
- The atomic number of Uranium is 92, and that of Lead is 82.
- The change in atomic number = Initial atomic number - Final atomic number
\[ \Delta Z = 92 - 82 = 10 \]

5. **Calculate the Number of Beta Particles Emitted**:
- Each beta particle (\( \beta \)) has a charge of -1 and does not change the mass number but decreases the atomic number by 1.
- Let \( b \) be the number of beta particles emitted.
- The equation for the atomic number becomes:
\[ 92 - 2a + b = 82 \]
- Substituting \( a = 7 \):
\[ 92 - 2(7) + b = 82 \]
\[ 92 - 14 + b = 82 \implies b = 82 - 78 = 4 \]

6. **Total Particles Emitted**:
- The total number of particles emitted is the sum of alpha and beta particles:
\[ \text{Total particles} = a + b = 7 + 4 = 11 \]

### Final Answer:
The total number of particles emitted in the decay process is **11**.

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