Consider two decay reactions.
(a) `_92^236Urarr_82^206Pb+10 protons+20 neutrons` (b) `_92^238Urarr_83^206Pb+8_2^4He+6 electrons`
Are both the reactions possible?

Explanation

To determine if both decay reactions are possible, we need to apply the laws of conservation of atomic number and mass number to each reaction.

### Step-by-Step Solution:

**(a)** For the reaction:
\[ _{92}^{236}U \rightarrow _{82}^{206}Pb + 10 \, \text{protons} + 20 \, \text{neutrons} \]

1. **Identify Atomic Numbers:**
- Left-hand side (LHS): Atomic number of Uranium (U) = 92
- Right-hand side (RHS): Atomic number of Lead (Pb) = 82 + (10 protons × 1) = 82 + 10 = 92

**Check Conservation of Atomic Number:**
- LHS = 92
- RHS = 92
- The atomic number is conserved.

2. **Identify Mass Numbers:**
- LHS: Mass number of Uranium = 236
- RHS: Mass number of Lead = 206 + (10 protons × 1) + (20 neutrons × 1) = 206 + 10 + 20 = 236

**Check Conservation of Mass Number:**
- LHS = 236
- RHS = 236
- The mass number is conserved.

**Conclusion for (a):** The reaction is possible.

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**(b)** For the reaction:
\[ _{92}^{238}U \rightarrow _{83}^{206}Pb + _{2}^{4}He + 6 \, \text{electrons} \]

1. **Identify Atomic Numbers:**
- LHS: Atomic number of Uranium (U) = 92
- RHS: Atomic number of Lead (Pb) = 83 + (2 for Helium) + (6 electrons × -1) = 83 + 2 - 6 = 79

**Check Conservation of Atomic Number:**
- LHS = 92
- RHS = 83 + 2 - 6 = 79
- The atomic number is not conserved.

2. **Identify Mass Numbers:**
- LHS: Mass number of Uranium = 238
- RHS: Mass number of Lead = 206 + (4 for Helium) + (0 for electrons) = 206 + 4 = 210

**Check Conservation of Mass Number:**
- LHS = 238
- RHS = 210
- The mass number is not conserved.

**Conclusion for (b):** The reaction is not possible.

### Final Conclusion:
- Reaction (a) is possible.
- Reaction (b) is not possible.