To solve the problem, we need to understand the concepts of isodiaphers and isobars, and how they relate to alpha and beta emissions.
### Step-by-Step Solution:
1. **Identify Isodiaphers:**
- Isodiaphers are nuclides that have the same value of \( N - Z \) (where \( N \) is the number of neutrons and \( Z \) is the number of protons).
- Given nuclides A, B, and C are isodiaphers.
2. **Determine the Mass and Atomic Numbers of A, B, and C:**
- Start with nuclide A, which is given as \( _{82}^{206}A \).
- This means A has:
- Mass number (A) = 206
- Atomic number (Z) = 82
- Neutrons (N) = Mass number - Atomic number = 206 - 82 = 124
3. **Alpha Emission from A to B:**
- In alpha emission, the mass number decreases by 4 and the atomic number decreases by 2.
- Therefore, for nuclide B:
- Mass number = 206 - 4 = 202
- Atomic number = 82 - 2 = 80
- So, \( B = _{80}^{202}B \).
4. **Alpha Emission from B to C:**
- Again, applying alpha emission:
- For nuclide C:
- Mass number = 202 - 4 = 198
- Atomic number = 80 - 2 = 78
- So, \( C = _{78}^{198}C \).
5. **Identify Isobars:**
- Isobars are nuclides that have the same mass number but different atomic numbers.
- Given nuclides C, D, and E are isobars.
6. **Beta Emission from C to D:**
- In beta emission, the mass number remains the same, but the atomic number increases by 1.
- Therefore, for nuclide D:
- Mass number = 198 (remains the same)
- Atomic number = 78 + 1 = 79
- So, \( D = _{79}^{198}D \).
7. **Beta Emission from D to E:**
- Again, applying beta emission:
- For nuclide E:
- Mass number = 198 (remains the same)
- Atomic number = 79 + 1 = 80
- So, \( E = _{80}^{198}E \).
8. **Calculate the Difference in Protons between A and E:**
- The atomic number of A is 82 (which is the number of protons).
- The atomic number of E is 80 (which is the number of protons).
- The difference in the number of protons between A and E is:
\[
\text{Difference} = Z_A - Z_E = 82 - 80 = 2
\]
### Final Answer:
The difference in the number of protons between A and E is **2**.
